{"id":13081,"date":"2021-03-26T00:33:45","date_gmt":"2021-03-26T00:33:45","guid":{"rendered":"http:\/\/Enoteprime.com\/?p=13081"},"modified":"2021-04-13T00:09:53","modified_gmt":"2021-04-13T00:09:53","slug":"Enote-on-mathematics-ss3-second-term","status":"publish","type":"post","link":"https:\/\/enoteprime.com\/Enote-on-mathematics-ss3-second-term\/","title":{"rendered":"Enote on Mathematics SS3 Second Term"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\" id=\"h-mathematics-topics-for-senior-secondary-school-Enoteprime-com\">Mathematics Topics for Senior Secondary School  &#8211;<a href=\"https:\/\/enoteprime.com\" target=\"_blank\" rel=\"noreferrer noopener\"> Enoteprime.com<\/a><\/h2>\n\n\n\n<p><\/p>\n\n\n\n<p><strong>SUBJECT:<\/strong> <strong>MATHEMATICS <strong>SECOND TERM<\/strong>&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>CLASS: SS 3<\/strong><\/p>\n\n\n\n<p><strong>SCHEME OF WORK<\/strong><\/p>\n\n\n\n<p><strong>WEEKS&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; TOPIC<\/strong>S<\/p>\n\n\n\n<ol type=\"1\"><li><strong>Calculation on interest on bonds and debentures using logarithm table and problems on taxes and value added tax.<\/strong><\/li><li><strong>Coordinate Geometry of straight line: Cartesian coordinate graphs, distance between two points, midpoint of the line joining two points.<\/strong><\/li><li><strong>Coordinate Geometry of straight lines: Gradient and Intercepts of a line, angle between two intersecting straight lines and application.<\/strong><\/li><li><strong>Differentiation of algebraic functions: meaning of differentiation, differentiation from first principle and standard derivatives of some basic functions.<\/strong><\/li><li><strong>Differentiation of algebraic functions: Basic rules of differentiation such as sum and difference, product rule, quotient rule and maximal and minimum application.<\/strong><\/li><li><strong>Integration and evaluation of simple algebraic functions: Definition, method of integration: substitution, partial fraction and integration by parts, area under the curve and use of Simpson\u2019s rule.<\/strong><\/li><li><strong>\u2013 12.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Revision and mock examination.<\/strong><\/li><\/ol>\n\n\n\n<p><strong>REFERENCE TEXT<\/strong><\/p>\n\n\n\n<ul><li><strong>New General Mathematics for SS book 3 by J.B Channon<\/strong><\/li><li><strong>Essential Mathematics for SS book 3<\/strong><\/li><li><strong>Mathematics Exam Focus<\/strong><\/li><li><strong>Waec and Jamb past Questions<\/strong><\/li><\/ul>\n\n\n\n<p><strong>WEEK ONE<\/strong><strong><\/strong><\/p>\n\n\n\n<ul><li>Calculation on interest on bonds and debentures using logarithm table<\/li><li>Problems on taxes and value added tax.<\/li><\/ul>\n\n\n\n<p><strong><br><\/strong><\/p>\n\n\n\n<p><strong>WEEK TWO<\/strong><strong><\/strong><\/p>\n\n\n\n<ul><li>Coordinate Geometry of straight line: Cartesian coordinate graphs<\/li><li>distance between two points<\/li><li>midpoint of the line joining two points<\/li><li>Coordinate Geometry of Straight line:<\/li><li>Cartesian coordinate graph:<\/li><\/ul>\n\n\n\n<p><strong>Distance between two lines:<\/strong><\/p>\n\n\n\n<p><strong>I<\/strong>n the figure below, the coordinates of the points A and B are (x<sub>1<\/sub>, y<sub>1<\/sub>) and (x<sub>2<\/sub>, y<sub>2<\/sub>), respectively. Let the length of AB be l.<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"12\" height=\"153\">y<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"126\" height=\"96\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B(x<sub>2<\/sub>, y<sub>2<\/sub>)<\/p>\n\n\n\n<p>l<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y<sub>2<\/sub> \u2013 y<sub>1<\/sub><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; A(x<sub>1<\/sub>, y<sub>1<\/sub>)&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;x<sub>2<\/sub> \u2013 x<sub>2 <\/sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"220\" height=\"12\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; X<\/p>\n\n\n\n<p>Using Pythagoras theorem:<\/p>\n\n\n\n<p>AB<sup>2<\/sup> = AC<sup>2<\/sup> + BC<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;l<sup>2<\/sup> =(x<sub>2<\/sub> \u2013 x<sub>1<\/sub>)<sup>2<\/sup> + (y<sub>2<\/sub> \u2013 y<sub>1<\/sub>)<sup>2<\/sup><\/p>\n\n\n\n<p><strong>l&nbsp; = \u221a(x<sub>2<\/sub> \u2013 x<sub>1<\/sub>)<sup>2<\/sup> + (y<sub>2<\/sub> \u2013 y<sub>1<\/sub>)<sup>2<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Example:<\/strong><\/p>\n\n\n\n<p>Find the distance between the each pair of points: a. (3, 4) and (1, 2)&nbsp;&nbsp; b. (3, &#8211; 3) and (- 2, 5)<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Using l =\u221a(x<sub>2<\/sub> \u2013 x<sub>1<\/sub>)<sup>2<\/sup> + (y<sub>2<\/sub> \u2013 y<sub>1<\/sub>)<sup>2<\/sup><\/p>\n\n\n\n<ol type=\"a\"><li>l = \u221a(3 \u2013 1)<sup>2<\/sup> + (4 \u2013 2)<sup>2<\/sup><\/li><\/ol>\n\n\n\n<p>l = \u221a2<sup>2<\/sup> + 2<sup>2<\/sup><\/p>\n\n\n\n<p>l = \u221a8 = 2\u221a2 units<\/p>\n\n\n\n<ul><li>l = \u221a(3 \u2013 (-2)<sup>2<\/sup> + (-3 \u2013 5)<sup>2<\/sup><\/li><\/ul>\n\n\n\n<p>=&nbsp; \u221a5<sup>2<\/sup> + (-8)<sup>2<\/sup><\/p>\n\n\n\n<p>=&nbsp; \u221a25 + 64 = \u221a89 = 9.43 units<\/p>\n\n\n\n<p><strong>Evaluation<\/strong>: Find the distance between the points in each of the following pairs leaving your answers in surd form:&nbsp; 1. (-2, &#8211; 5) and (3, &#8211; 6)&nbsp;&nbsp;&nbsp;&nbsp; 2. (- 3, 4) and (- 1, 2)<\/p>\n\n\n\n<p><strong>Mid-point of a line:<\/strong><\/p>\n\n\n\n<p>The mid-point of the line joining two points:<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"12\" height=\"197\">y<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"188\" height=\"138\">&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B(x<sub>2<\/sub>, y<sub>2<\/sub>)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y<sub>2<\/sub> &#8211; y<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"2\" height=\"75\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"88\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;M(x, y)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D&nbsp;&nbsp;<\/p>\n\n\n\n<p>x<sub>2<\/sub> \u2013x&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y \u2013 y<sub>1<\/sub><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; A(x<sub>1<\/sub>, y<sub>1<\/sub>)&nbsp;&nbsp; x \u2013 x<sub>1 <\/sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;N&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; C<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"303\" height=\"12\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; X<\/p>\n\n\n\n<p>&nbsp;Triangle MAN and BMD are congruent, so AM = MD and BD = MN<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x \u2013 x<sub>1<\/sub> = x<sub>2<\/sub> \u2013 x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y \u2013 y<sub>1<\/sub> = y<sub>2<\/sub> \u2013 y<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x + x = x<sub>2<\/sub> + x<sub>1<\/sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y + y = y<sub>2<\/sub> + y<sub>1<\/sub><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2x = x<sub>2<\/sub> + x<sub>1<\/sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2y&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = y<sub>2<\/sub> + y<sub>1<\/sub><\/p>\n\n\n\n<p><strong>x= <u>x<sub>2<\/sub> + x<sub>1<\/sub><\/u>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = <u>y<sub>2<\/sub> + y<sub>1<\/sub><\/u><\/strong><\/p>\n\n\n\n<p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2<\/strong><\/p>\n\n\n\n<p><strong><\/strong><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"126\" height=\"35\">Hence, the <strong>mid-point<\/strong> of a straight line joining two is&nbsp;&nbsp;&nbsp; <strong><u>x<sub>2<\/sub> + x<sub>1<\/sub><\/u>&nbsp; ,<u>y<sub>2<\/sub> + y<sub>1<\/sub><\/u><\/strong><\/p>\n\n\n\n<p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2<\/strong><\/p>\n\n\n\n<p><strong>Example<\/strong>: Find the coordinates of the mid-point of the line joining the following pairs of points.<\/p>\n\n\n\n<ol type=\"a\"><li>(3, 4) and (1, 2)&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b. (2, 5) and ( &#8211; 3, 6)<\/li><\/ol>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"126\" height=\"35\">Solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp; Mid-point =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong><u>x<sub>2<\/sub> + x<sub>1<\/sub><\/u>&nbsp; ,<u>y<sub>2<\/sub> + y<sub>1<\/sub><\/u><\/strong><\/p>\n\n\n\n<p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2<\/strong><\/p>\n\n\n\n<ol type=\"a\"><li><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"106\" height=\"35\">Mid-point =&nbsp; 1 + 3&nbsp; ,&nbsp;&nbsp;&nbsp;&nbsp; 4 + 2&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=&nbsp; (2, 3)<\/li><\/ol>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2<\/p>\n\n\n\n<ul><li><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"106\" height=\"35\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"68\" height=\"35\">Mid-point = &nbsp;&#8211; 3 + 2 &nbsp;,&nbsp; &nbsp;6+ 5 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=&nbsp;&nbsp; &#8211; 1&nbsp; ,&nbsp;&nbsp; 11<\/li><\/ul>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2<\/p>\n\n\n\n<p><strong>Evaluation:<\/strong> Find the coordinates of the mid-point of the line joining the following pairs of points.<\/p>\n\n\n\n<ol type=\"a\"><li>(- 2 , &#8211; 5) and (3, &#8211; 6)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; b. (3, 4) and ( &#8211; 1, &#8211; 2)<\/li><\/ol>\n\n\n\n<p><strong>General Evaluation<\/strong><\/p>\n\n\n\n<ol type=\"1\"><li>Find the distance between the points in each of the following pairs leaving your answers&nbsp;&nbsp;&nbsp;&nbsp; in surd form:&nbsp; 1. (7, 2) and (1, 6)<\/li><li>What is the value of r if the distance between the points (4, 2) and (1, r) is 3 units?<\/li><li>Find the coordinates of the mid-point (-3, -2) and (-7, &#8211; 4)<\/li><\/ol>\n\n\n\n<p><strong>Reading Assignment: NGM for SS 3 Chapter 9 <\/strong>page 77 \u2013 78,<\/p>\n\n\n\n<p><strong>Weekend Assignment:<\/strong><\/p>\n\n\n\n<ol type=\"1\"><li>Find the value of \u03b1<sup>2<\/sup> + \u03b2<sup>2<\/sup> if \u03b1 + \u03b2 = 2 and the distance between the points (1, \u03b1) and (\u03b2, 1) is 3 units.<\/li><li>The vertices of the triangle ABC are A (7, 7), B (- 4, 3) and C (2, &#8211; 5). Calculate the length of the longest side of triangle ABC.<\/li><li>Using the information in \u20182\u2019 above, calculate the line AM, where M is the mid-point of the side opposite A.<\/li><\/ol>\n\n\n\n<p><strong>WEEK THREE<\/strong><strong><\/strong><\/p>\n\n\n\n<ul><li>Coordinate Geometry of straight lines:<\/li><li>Gradient and Intercepts of a line<\/li><li>Angle between two intersecting straight lines and application<\/li><\/ul>\n\n\n\n<p><strong>Gradient and Intercepts of a line<\/strong><\/p>\n\n\n\n<p>Gradient of a line of the form y = mx + c, is the coefficient of x, which is represented by m and c is the intercept on the y axis.<\/p>\n\n\n\n<p><strong>Example<\/strong><\/p>\n\n\n\n<p>1. Find the equation of the line with gradient 4 and y-intercept -7.<\/p>\n\n\n\n<p><strong>Solution<\/strong><\/p>\n\n\n\n<p>m = 4, c = &#8211; 7,<\/p>\n\n\n\n<p>Hence, the equation is; y =4x &#8211; 7.<\/p>\n\n\n\n<p><strong>Evaluation:<\/strong><\/p>\n\n\n\n<p>1.What is the gradient and y intercept of the line equation 3x -5y +10=0 ?<\/p>\n\n\n\n<p>2. Find the equation of the line with gradient &#8211; 9 and y-intercept 4.<\/p>\n\n\n\n<p><strong>Gradient and One Point Form<\/strong><\/p>\n\n\n\n<p>The equation of the line can be calculated given one point (x, y) and gradient (m) by using the formula; y &#8211; y1= m(x &#8211; x1)<\/p>\n\n\n\n<p><strong>Example<\/strong><\/p>\n\n\n\n<p>Find the equation of the line with gradient -8 and point(3, 7).<\/p>\n\n\n\n<p><strong>Solution<\/strong><\/p>\n\n\n\n<p>m = &#8211; 8, (x1, y1) =(3,7)<\/p>\n\n\n\n<p>Equation: y &#8211; 7 = &#8211; 8(x &#8211; 3)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = -8x + 24 +7<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = -8x + 31<\/p>\n\n\n\n<p><strong>Evaluation:<\/strong><\/p>\n\n\n\n<p>1. Find the equation of the line with gradient 5 and point(-2, -7).<\/p>\n\n\n\n<p>2. Find the equation of the line with gradient -12and point (3, -5).<\/p>\n\n\n\n<p><strong>Two Point Form:<\/strong><\/p>\n\n\n\n<p>Given two points (x1, y1) and (x2, y2), the equation can be obtained using the formula:<\/p>\n\n\n\n<p>&nbsp;y2 &#8211; y1 = y &#8211; y1<\/p>\n\n\n\n<p>x2 &#8211; x1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x &#8211; x1<\/p>\n\n\n\n<p>Example: Find the equation of the line passing through (2,-5) and (3,6).<\/p>\n\n\n\n<p><strong>Solution<\/strong><\/p>\n\n\n\n<p>6 &#8211; (-5)\/3 &#8211; 2 = y &#8211; (-5)\/x &#8211; 2<\/p>\n\n\n\n<p>11 = y + 5\/x &#8211; 2<\/p>\n\n\n\n<p>11(x &#8211; 2) = y + 5<\/p>\n\n\n\n<p>11x &#8211; 22 = y + 5<\/p>\n\n\n\n<p>y &#8211; 11x + 27 = 0<\/p>\n\n\n\n<p><strong>Evaluation:<\/strong><\/p>\n\n\n\n<p>1.Find the equation of the line passing through (3, 4) and (-1, -2).<\/p>\n\n\n\n<p>2.Find the equation of the line passing through (-8, 5) and (-6, 2).<\/p>\n\n\n\n<p><strong>Angles between Lines<\/strong><\/p>\n\n\n\n<p><strong>Parallel lines:<\/strong><\/p>\n\n\n\n<p>The angle between parallel lines is 0<sup>0<\/sup> because they have the same gradient<\/p>\n\n\n\n<p><strong>Perpendicular Lines:<\/strong><\/p>\n\n\n\n<p>Angle between two perpendicular lines is 90<sup>0<\/sup> and the product of their gradients is \u2013 1. Hence, m<sub>1<\/sub>m<sub>2<\/sub> = &#8211; 1<\/p>\n\n\n\n<p><strong>Examples:<\/strong><\/p>\n\n\n\n<ol type=\"1\"><li>Show that the lines y = -3x + 2 and y + 3x = 7 are parallel.<\/li><\/ol>\n\n\n\n<p>solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Equation 1: y = -3x + 2,&nbsp;&nbsp; m<sub>1<\/sub> = -3<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Equation 2:&nbsp; y + 3x = 7,&nbsp;&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = -3x + 7, m<sub>2<\/sub> = &#8211; 3<\/p>\n\n\n\n<p>since; m<sub>1<\/sub> = m<sub>2<\/sub> = &#8211; 3, then the lines are parallel<\/p>\n\n\n\n<ul><li>Given the line equations x = 3y + 5 and y + 3x = 2, show that the lines are perpendicular.<\/li><\/ul>\n\n\n\n<p>solutions:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp; Equation 1:&nbsp;&nbsp;&nbsp;&nbsp; x = 3y + 5,&nbsp;&nbsp; make y the subject of the equation.<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3y = x + 5<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = x\/3 + 5\/3<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; m<sub>1<\/sub> = 1\/3<\/p>\n\n\n\n<p>&nbsp; Equation 2:&nbsp;&nbsp;&nbsp; y + 3x = 2,<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = &#8211; 3x + 2,&nbsp;&nbsp; m<sub>2<\/sub> = -3<\/p>\n\n\n\n<p>hence: m<sub>1<\/sub> x m<sub>2<\/sub> = 1\/3 x \u2013 3 = &#8211; 1<\/p>\n\n\n\n<p>since: m<sub>1<\/sub>m<sub>2<\/sub> = &#8211; 1, then the lines are perpendicular.<\/p>\n\n\n\n<p><strong>Evaluation: <\/strong>State which of the following pairs of lines are: (i) perpendicular&nbsp;&nbsp; (ii) parallel<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (1)&nbsp;&nbsp;&nbsp; y = x + 5 and y = &#8211; x + 5&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;(2). 2y \u2013 6 = 5x and 3 \u2013 5y = 2x&nbsp;&nbsp;&nbsp; (3) y = 2x \u2013 1 and 2y \u2013 4x = 8 <strong><br><\/strong><\/p>\n\n\n\n<p><strong>Angles between Intersecting Lines:<\/strong><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"12\" height=\"156\"><strong>y<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"219\" height=\"97\"><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = mx + c&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"222\" height=\"12\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u03b8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 0&nbsp;&nbsp;<\/p>\n\n\n\n<p>The gradient of y = mx + c is tan \u03b8.&nbsp;&nbsp;&nbsp; Hence <strong>m = tan \u03b8<\/strong>can be used to calculate angles between two intersecting lines. Generally the angle between two lines can be obtained using: tan 0 = m2 -m1<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1 + m1m2<\/p>\n\n\n\n<p>Example: Calculate the acute angle between the lines y=4x -7 and y = x\/2 + 0.5.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Y=4x -7, m1= 4, y=x\/2+0.2, m2 =1\/2.<\/p>\n\n\n\n<p>&nbsp;Tan O= 0.5 &#8211; 4.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = -3.5\/3<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1 + (0.5*4)<\/p>\n\n\n\n<p>Tan O =- 1.1667<\/p>\n\n\n\n<p>O=tan-1(-1.1667) = 49.4<\/p>\n\n\n\n<p>Evaluation:Calculate the acute angle between the lines y=3x -4 and x &#8211; 4y +8 = 0.<\/p>\n\n\n\n<p>General Evaluation:<\/p>\n\n\n\n<p>1.Calculate the acute angle between the lines y=2x -1 and&nbsp; 2y + x = 2.<\/p>\n\n\n\n<p>2.If the lines 3y=4x -1 and qy= x + 3 are parallel to each other, find the value of q.<\/p>\n\n\n\n<p>3.Find the equation of the line passing through (2,-1) and gradient 3.<\/p>\n\n\n\n<p><strong>Reading Assignment: NGM for SS 3 Chapter 9 <\/strong>page 75 \u2013 81<\/p>\n\n\n\n<p>Weekend Assignment<\/p>\n\n\n\n<p>1.Find the equation of the line passing through (5,0) and gradient 3.<\/p>\n\n\n\n<p>2.Find the equation of the line passing through (2,-1) and (1, -2).<\/p>\n\n\n\n<p>&nbsp;3. Two lines y=3x &#8211; 4 and x &#8211; 4y + 8=0 are drawn on the same axes.<\/p>\n\n\n\n<p>(a) Find the gradients and intercepts on the axes of each line.<\/p>\n\n\n\n<p>(b) Find the equation parralel to x -4y + 8=0 at the point (3, -5)<\/p>\n\n\n\n<p><\/p>\n\n\n\n<p>Mathematics Topics for Senior Secondary School &#8211;<a href=\"https:\/\/enoteprime.com\" target=\"_blank\" rel=\"noreferrer noopener\"> Enoteprime.com<\/a><\/p>\n\n\n\n<p><strong>WEEK FOUR<\/strong><strong><\/strong><\/p>\n\n\n\n<ul><li>Differentiation of algebraic functions: meaning of differentiation<\/li><li>Differentiation from first principle<\/li><li>Standard derivatives of some basic functions.<\/li><\/ul>\n\n\n\n<p>Consider the curve whose equation is given by&nbsp;&nbsp; y = f(x)&nbsp; Recall that m = y<sub>2<\/sub> \u2013 y<sub>1<\/sub>= f(x+x)-f(x)<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"77\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"45\" height=\"2\">x<sub>2<\/sub>&#8211; x<sub>1<\/sub>x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>As point B moves close to A, dx becomes smaller and tends to zero.<\/p>\n\n\n\n<p>The limiting value is written on&nbsp; Lim&nbsp;&nbsp; f(x +x) \u2013 f(x) and is denoted by as x \u2013&gt; 0<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"92\" height=\"3\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dx<\/p>\n\n\n\n<p>f<sup>l<\/sup>(x) is called the <strong>derivative of f(x)<\/strong> and the <strong>gradient function of the curve<\/strong><\/p>\n\n\n\n<p>The process of finding the derivative of f(x) is called differentiation. The rotations which are commonly used for the derivative of a function are f<sup>1<\/sup>(x) read as f \u2013 prime of x,&nbsp; df\/dx read as&nbsp; dee x of f<\/p>\n\n\n\n<p>df\/dx&nbsp;&nbsp;&nbsp; read&nbsp; dee &#8211; f&nbsp; dee- x,&nbsp;&nbsp;&nbsp; dy\/dx read&nbsp; dee &#8211; y&nbsp; dee- x<\/p>\n\n\n\n<p><strong>If&nbsp;&nbsp; y = f(x) , this dy\/dx = f<sup>1<\/sup>(x) (it is called the differential coefficient of y with respect to x.<\/strong><\/p>\n\n\n\n<p><strong>Differentiation from first principle:<\/strong> The process of finding the derivative of a function from the consideration of the limiting value is called differentiation from first principle.<\/p>\n\n\n\n<p><strong>Example 1<\/strong><\/p>\n\n\n\n<p>Find from first principle, the derivative of&nbsp;&nbsp; y = x<sup>2<\/sup><\/p>\n\n\n\n<p>Solution<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = x<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y + y = (x + x)<sup>2<\/sup><\/p>\n\n\n\n<p>y + y = x<sup>2<\/sup> + 2xx + (x)<sup>2<\/sup><\/p>\n\n\n\n<p>y =&nbsp; x<sup>2<\/sup> + 2xx+ (x)<sup>2 <\/sup>&nbsp;&#8211;&nbsp; y&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>y = x<sup>2<\/sup>&nbsp; +&nbsp; 2xx&nbsp;&nbsp; +&nbsp; (x)<sup>2<\/sup>&nbsp; &#8211;&nbsp; x<sup>2<\/sup><\/p>\n\n\n\n<p>y&nbsp; =&nbsp; 2xx&nbsp; +&nbsp; (x)<sup>2<\/sup><\/p>\n\n\n\n<p>y&nbsp; = (2x&nbsp;&nbsp; +&nbsp; x)x<\/p>\n\n\n\n<p>y =&nbsp; 2x&nbsp;&nbsp; +&nbsp;&nbsp; x<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"17\" height=\"2\">x<\/p>\n\n\n\n<p>Lim&nbsp; x&nbsp; =&nbsp; 0<\/p>\n\n\n\n<p>dy =&nbsp; 2x<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"18\" height=\"2\">dx<\/p>\n\n\n\n<p><strong>Example 2<\/strong>:<\/p>\n\n\n\n<p>Find from first principle, the derivative of 1\/x<\/p>\n\n\n\n<p>Solution<\/p>\n\n\n\n<p>Let y&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"14\" height=\"2\">x<\/p>\n\n\n\n<p>y + y =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"33\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x + x<\/p>\n\n\n\n<p>y&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &#8211;&nbsp; y<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"41\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x + x<\/p>\n\n\n\n<p>y =&nbsp;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &#8211;&nbsp; 1<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"25\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"42\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x + x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<\/p>\n\n\n\n<p>y&nbsp; =&nbsp; x \u2013 (x +&nbsp; x)<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"74\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (x&nbsp; +x)x<\/p>\n\n\n\n<p>y =&nbsp; x&nbsp; &#8211;&nbsp; x&nbsp; &#8211;&nbsp; x<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"69\" height=\"2\">x<sup>2<\/sup>&nbsp; +&nbsp; xx<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"43\" height=\"2\">dy&nbsp; =&nbsp;&nbsp;&nbsp; -x<\/p>\n\n\n\n<p>x<sup>2<\/sup>+ x<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"17\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"42\" height=\"2\">y&nbsp; =&nbsp;&nbsp; -1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>x&nbsp;&nbsp;&nbsp;&nbsp; x<sup>2<\/sup> + x&nbsp;&nbsp;<\/p>\n\n\n\n<p>Lim&nbsp; x = 0<\/p>\n\n\n\n<p>dy =&nbsp;&nbsp;&nbsp; -1<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"19\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"19\" height=\"2\">dx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<sup>2<\/sup><\/p>\n\n\n\n<p><strong>Evaluation: <\/strong>Find from first principle, the derivatives of y with respect to x:<\/p>\n\n\n\n<ol type=\"1\"><li>Y = 3x<sup>3<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2. Y = 7x<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp; 3. Y = 3x<sup>2<\/sup> \u2013 5x<\/li><\/ol>\n\n\n\n<p><strong>Rules of Differentiation:&nbsp;&nbsp;&nbsp;&nbsp; <\/strong>Let&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = x<sup>n<\/sup><\/p>\n\n\n\n<p>y + dy = (x + dx)<sup>n<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = x<sup>n<\/sup> + nx<sup>n-1<\/sup>dx + n(n -1) x<sup>n-2<\/sup>(dx)<sup>2<\/sup> + \u2026 (dx)<sup>n<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"98\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; 2!<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = x<sup>n<\/sup> + n x<sup>n-1<\/sup>dx + n(n-1) x<sup>n-2&nbsp; <\/sup>(dx)<sup>2<\/sup>+ &#8212; + (dx)<sup>n<\/sup> &#8211; x<sup>n<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"101\" height=\"3\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2!<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = nx<sup>n-1<\/sup>dx + n (n \u2013 1) x<sup>n\u20131 <\/sup>(dx)<sup>2<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"114\" height=\"3\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2!<\/p>\n\n\n\n<p>dy\/dx = n <sup>xn-1<\/sup> + n (n \u20131) x<sup>n-1 <\/sup>dx<\/p>\n\n\n\n<p>Lim dy\/dx = nx<sup>n-1<\/sup><\/p>\n\n\n\n<p>dx = 0<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hence;&nbsp;&nbsp; <strong>dy\/dx&nbsp; =&nbsp; nx<sup>n-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/sup>if y = x<sup>n<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Example 3<\/strong>:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>Find the derivative of the following with respect to x:&nbsp;&nbsp; (a) x<sup>7<\/sup> (b) x<sup>\u00bd<\/sup> (c) 5x<sup>2<\/sup> \u2013 3x (d) &#8211; 3x<sup>2<\/sup> (e) y = 2x<sup>3<\/sup> \u2013 3x + 8<\/p>\n\n\n\n<p>Solution<\/p>\n\n\n\n<p>a.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let&nbsp; y = x<sup>7<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = 7 x<sup>7-1 <\/sup>= 7x<sup>6<\/sup><\/p>\n\n\n\n<p>b.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let&nbsp; y = x <sup>\u00bd<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"35\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = \u00bd x<sup>\u00bd -1 <\/sup>= \u00bd x<sup>\u2013 \u00bd<\/sup>&nbsp; =&nbsp;&nbsp; 1<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2\u221ax<\/p>\n\n\n\n<p>c.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = 5x<sup>2<\/sup> \u2013 3x<\/p>\n\n\n\n<p>dy\/dx = 10x \u2013 3<\/p>\n\n\n\n<p>d.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = &#8211; 3x<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx =2\u00d7 &#8211; 3x<sup>2-1<\/sup> = &#8211; 6x<\/p>\n\n\n\n<p>&nbsp;e.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = 2x<sup>3<\/sup> \u2013 3x + 8<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx= 3 x 2x<sup>3-1 <\/sup>\u2013 3 + 0<\/p>\n\n\n\n<p>= 6x<sup>2<\/sup> \u2013 3<\/p>\n\n\n\n<p><strong>Evaluation:<\/strong><\/p>\n\n\n\n<ol type=\"1\"><li>If&nbsp; y=5x<sup>4<\/sup> ,find&nbsp; dy\/dx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2.Given that y= 4x<sup>-1<\/sup> find y<sup>1<\/sup><\/li><\/ol>\n\n\n\n<p><strong>General Evaluation<\/strong><\/p>\n\n\n\n<p>&nbsp;1. Find, from first principles, the derivative of&nbsp; 4x<sup>2<\/sup> \u2013 2&nbsp; with&nbsp; respect&nbsp; to&nbsp; x.<\/p>\n\n\n\n<p>&nbsp;2. Find the derivative of the following&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a.3x<sup>3 <\/sup>\u2013 7x<sup>2<\/sup> \u2013 9x + 4&nbsp;&nbsp; b. 2x<sup>3<\/sup>&nbsp;&nbsp;&nbsp; c. 3\/x<\/p>\n\n\n\n<p><strong>3. <\/strong>Using idea of difference of two square; simplify 243x<sup>2<\/sup> &#8211; 48y<sup>2<\/sup><\/p>\n\n\n\n<p>4. Expand (2x -5)( 3x-4)<\/p>\n\n\n\n<p>5. If the gradient of y=2x<sup>2<\/sup>-5 is -12 find the value of y.<\/p>\n\n\n\n<p><strong>Reading Assignment: NGM for SS 3 Chapter 10 <\/strong>page 82 -88,<\/p>\n\n\n\n<p><strong>Weekend Assignment<\/strong><\/p>\n\n\n\n<p><strong>Objective<\/strong><\/p>\n\n\n\n<p>1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find the derivative of 5x<sup>3<\/sup>(a) 10x<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp; (b) 15x<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (c) 10x&nbsp;&nbsp;&nbsp;&nbsp; (d) 15x<sup>3<\/sup><\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find dy\/dx, if y = 1\/x<sup>3<\/sup>(a) \u20133\/x<sup>4<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (b) 3\/x<sup>4 <\/sup>&nbsp;&nbsp;&nbsp;&nbsp;(c) 4\/x<sup>3<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (e) \u20134\/x<sup>3<\/sup><\/p>\n\n\n\n<p>3.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find f<sup>1<\/sup>(x), if f(x) = x<sup>3<\/sup> (a) 3x &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (b) 3x<sup>2<\/sup> (c) \u00bd x<sup>3<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (d) 2x<sup>3<\/sup><\/p>\n\n\n\n<p>4.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find the derivative of&nbsp;&nbsp; 1\/x(a) 1\/x<sup>2<\/sup>&nbsp;&nbsp;&nbsp; (b) \u20131\/x<sup>2<\/sup>&nbsp;&nbsp;&nbsp; (c) \u2013 x &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (d) \u2013x<sup>2<\/sup><\/p>\n\n\n\n<p>5.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; If&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = &#8211; 2\/3 x<sup>3<\/sup>. Find dy\/dx (a) 4\/3 x<sup>2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/sup>&nbsp;(b) 2x<sup>2<\/sup>&nbsp;&nbsp;&nbsp; (c) \u2013 2x<sup>2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/sup>(d) \u20132x<\/p>\n\n\n\n<p><strong><u><br><\/u><\/strong><strong>Theory<\/strong><strong><u><\/u><\/strong><\/p>\n\n\n\n<p>1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find from first principle, the derivative of&nbsp;&nbsp; y = x + 1\/x<\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find the derivative of 2x<sup>2<\/sup> \u2013 2\/x<sup>3<\/sup><\/p>\n\n\n\n<p><strong>WEEK FIVE<\/strong><strong><\/strong><\/p>\n\n\n\n<ul><li>Differentiation of algebraic functions:<\/li><li>Basic rules of differentiation such as sum and difference, product rule, quotient rule<\/li><li>Maximal and minimum application.<\/li><\/ul>\n\n\n\n<p><strong>Derivative of algebraic functions<\/strong><\/p>\n\n\n\n<p>Let f, u, v be functions such that<\/p>\n\n\n\n<p>f(x) = u(x) + v(x)<\/p>\n\n\n\n<p>f(x +x ) = u(x +x ) + v(x + x)<\/p>\n\n\n\n<p>f(x + x) \u2013 f(x) = {u(x+ x) + v(x+ x) \u2013 v(x + x) \u2013 u(x) \u2013 v(x)}<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = u (x +x ) \u2013 u(x) + v(x +x ) \u2013 v(x)<\/p>\n\n\n\n<p>f(x + x) \u2013 f(x) = u(x +x)-u(x) + v(x +x ) \u2013 v(x)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td><\/td><td><\/td><td><\/td><td><\/td><td><\/td><td><\/td><\/tr><tr><td><\/td><td><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"87\" height=\"3\"><\/td><\/tr><tr><td><\/td><td><\/td><td><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"86\" height=\"2\"><\/td><td><\/td><td><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"99\" height=\"2\"><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Lim&nbsp; f(x + x) \u2013 f(x) = U<sup>1<\/sup>(x) + V<sup>1<\/sup>(x)&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>if y = u + v and u and v are functions of x, then <strong>dy\/dx = du\/dx + dv\/dx<\/strong><\/p>\n\n\n\n<p><strong><u>Examples<\/u><\/strong>:Find the derivative of the following<\/p>\n\n\n\n<p>1) 2x<sup>3<\/sup> \u2013 5 x<sup>2<\/sup> + 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2)3x<sup>2<\/sup> + 1\/x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3)2x<sup>3<\/sup> + 2x<sup>2<\/sup> +1<\/p>\n\n\n\n<p>Solution<\/p>\n\n\n\n<p>1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = 2x<sup>3<\/sup> \u2013 5x<sup>2<\/sup> + 2<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = 6x<sup>2<\/sup> \u2013 10x<\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = 3x<sup>2<\/sup>+ 1\/x = 3x<sup>2<\/sup> + x<sup>-1<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = 6x \u2013 x<sup>-2<\/sup> = 6x \u2013 1<\/p>\n\n\n\n<p>x<sup>2<\/sup><\/p>\n\n\n\n<p>3.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = 2x<sup>3<\/sup> + 2x<sup>2<\/sup> + 1<\/p>\n\n\n\n<p>dy\/dx=6x<sup>2<\/sup>&nbsp; +&nbsp; 4x<\/p>\n\n\n\n<p><strong>Evaluation:&nbsp;&nbsp; <\/strong>1. If y = 3x<sup>4<\/sup> \u2013 2x<sup>3<\/sup> \u2013 7x + 5. Find dy\/dx<\/p>\n\n\n\n<p>2.Findd (8x<sup>3<\/sup> \u2013 5x<sup>2<\/sup> + 6)<\/p>\n\n\n\n<p>Dx<\/p>\n\n\n\n<p><strong>Function of a function (chain rule)<\/strong><\/p>\n\n\n\n<p>Suppose that we know that y is a function of u and that u is a function of x, how do we find the derivative of y with respect to x?<\/p>\n\n\n\n<p>Given that y = f(x) and u = h(x), what is dy\/dx?<\/p>\n\n\n\n<p>dy\/dx = ,&nbsp;&nbsp; this is called the chain rule<\/p>\n\n\n\n<p><strong>Examples<\/strong><\/p>\n\n\n\n<p>Find the derivative of the following.(a)y = (3x<sup>2<\/sup> \u2013 2)<sup>3<\/sup>&nbsp;&nbsp; (b)&nbsp; y = (1 \u2013 2x<sup>3<\/sup>)&nbsp; (c) 5\/(6-x<sup>2<\/sup>)<sup>3<\/sup><\/p>\n\n\n\n<p>Solution<\/p>\n\n\n\n<p>1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = (3x<sup>2<\/sup> \u2013 2)<sup>3<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let u = 3x<sup>2<\/sup> \u2013 2<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = (3x<sup>2<\/sup> \u2013 2)<sup>3<\/sup> =&gt; y = u<sup>3<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = u<sup>3<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/du = 3u<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; du\/dx = 6x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx =&nbsp; = 3u<sup>2<\/sup> x 6x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 18xu<sup>2<\/sup> = 18x(3x<sup>2<\/sup> \u2013 2)<sup>2<\/sup><\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = (1 \u2013 2x<sup>3<\/sup>)<sup>1\/2<\/sup> =&gt; (1 \u2013 2x<sup>3<\/sup>) <sup>1\/2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let u =1 \u2013 2x<sup>3<\/sup>,&nbsp;&nbsp;&nbsp;&nbsp; hence&nbsp; y = u<sup>1\/2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = = \u00bd u<sup>-1\/2<\/sup>&nbsp; x( \u20136x<sup>2<\/sup>)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = -3x<sup>2<\/sup> u <sup>\u2013 \u00bd<\/sup>= &nbsp;-3x<sup>2<\/sup><\/p>\n\n\n\n<p>u<sup>1\/2<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"63\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"21\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; -3x<sup>2<\/sup>&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; -3x<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"63\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"21\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; \u221a u&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; \u221a(1 \u2013 2x<sup>3<\/sup>)<\/p>\n\n\n\n<p>3.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 5(6 \u2013 x<sup>2<\/sup>)<sup>-3<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"50\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (6 \u2013 x<sup>2<\/sup>)<sup>3<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let u = 6 \u2013 x<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = 5(u)<sup>\u20133<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/du = -15u <sup>\u20134<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; du\/dx = -2x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = dy\/du X du\/dx = -15u<sup>-4<\/sup> x (-2x) = 30x u<sup>-4<\/sup> = 30x (6 \u2013 x<sup>2<\/sup>)<sup>-4<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;30x_<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (6 \u2013 x<sup>2<\/sup>)<sup>4<\/sup><\/p>\n\n\n\n<p><strong>Evaluation<\/strong>:<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"48\" height=\"3\">1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Given that y =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; find dy\/dx<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (2x + 3)<sup>4<\/sup><\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; If y = (3x<sup>2<\/sup> + 1)<sup>3<\/sup> , Find dy\/dx<\/p>\n\n\n\n<p><strong>Product Rule<\/strong><\/p>\n\n\n\n<p>We shall consider the derivative of y = uv where u and v are function of x.<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let y = uv<\/p>\n\n\n\n<p>Then y + y = (u +u )(v + v)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = uv + uv + vu +uv<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = uv + uv + vu+ uv \u2013 uv<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y= uv + vu + uv<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y=&nbsp; uv&nbsp; +&nbsp;&nbsp;&nbsp; vu&nbsp;&nbsp; +&nbsp;&nbsp;&nbsp; uv<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; As x =&gt;0 ,u=&gt; 0 , v=&gt; 0<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"19\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Lim&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; Lim&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; uv +&nbsp;&nbsp;&nbsp; Lim&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; vu&nbsp;&nbsp; +&nbsp;&nbsp;&nbsp;&nbsp; Lim&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; uv<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"19\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"20\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"19\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=&gt;0&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=&gt;0&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=&gt;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=&gt;0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<\/p>\n\n\n\n<p>Hence&nbsp;&nbsp; <strong>dy\/dx<\/strong>= <strong>U dv + Vdu<\/strong><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"20\" height=\"2\"><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"21\" height=\"2\"><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dxdx<\/strong><\/p>\n\n\n\n<p><strong>Examples<\/strong><\/p>\n\n\n\n<p>Find the derivatives of the following.<\/p>\n\n\n\n<p>(a)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = (3 + 2x) (1 \u2013 x)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (b) y = (1 \u2013 2x + 3x<sup>2<\/sup>) (4 \u2013 5x<sup>2<\/sup>)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Solution<\/p>\n\n\n\n<p>1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = (3 + 2x) (1 \u2013 x)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let u = 3 + 2x and v = (1 \u2013 x)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; du\/dx = 2 and dv\/dx = -1<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dv\/dx = u dv + vdu<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"21\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dx<\/p>\n\n\n\n<p>=&nbsp; (1 \u2013 x) 2 + (3 + 2x) (-1)&nbsp;&nbsp;&nbsp;&nbsp; = 2 \u2013 2x \u2013 3 \u2013 2x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx&nbsp;&nbsp; = &#8211; 1 \u2013 4x<\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = (1 \u2013 2x + 3x<sup>2<\/sup>) (4 \u2013 5x<sup>2<\/sup>)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let u = (1 \u2013 2x + 3x<sup>2<\/sup>)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; and v = (4 \u2013 5x<sup>2<\/sup>)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; du\/dx = -2 + 6x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; and dv\/dx = &#8211; 10x<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"13\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dy\/dx = udv +&nbsp;&nbsp; vdu<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"17\" height=\"2\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dxdx<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = (1 \u2013 2x + 3x<sup>2<\/sup>) (-10x) + (4 \u2013 5x<sup>2<\/sup>) (- 2 + 6x)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &#8211; 10x + 20x<sup>2<\/sup> \u2013 30x<sup>3<\/sup> + (- 8 + 10x<sup>2<\/sup> + 24x \u2013 30x<sup>3<\/sup>)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &#8211; 10x + 20x<sup>2<\/sup> \u2013 30x<sup>3<\/sup> \u2013 8 + 10x<sup>2<\/sup> + 24x \u2013 30x<sup>3<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 14x + 30x<sup>2<\/sup> \u2013 60x<sup>3<\/sup> \u2013 8<\/p>\n\n\n\n<p><strong>Evaluation<\/strong><\/p>\n\n\n\n<p>Given that (i) y = (5+3x)(2-x)&nbsp;&nbsp;&nbsp; (ii)&nbsp; y = (1+x)(x+2)<sup>3\/2<\/sup><strong> ,<\/strong>find dy\/dx<\/p>\n\n\n\n<p><strong>Quotient Rule:<\/strong><\/p>\n\n\n\n<p>If y = u<\/p>\n\n\n\n<p><strong>v<\/strong><\/p>\n\n\n\n<p>then; <strong><u>dy<\/u><\/strong><strong> =&nbsp; v<u>du<\/u>&#8211; u<u>dv<\/u><\/strong><\/p>\n\n\n\n<p><strong>dxdxdx<\/strong><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"72\" height=\"3\"><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v<sup>2<\/sup><\/strong><\/p>\n\n\n\n<p><strong>E<\/strong>xamples:<\/p>\n\n\n\n<p>Differentiate the following with respect to x.&nbsp; (a)x<sup>2<\/sup> + 1&nbsp;&nbsp;&nbsp;&nbsp; (b)(x \u2013 1)<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1 \u2013 x<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u221ax<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol type=\"1\"><li>y = x<sup>2<\/sup> + 1<\/li><\/ol>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1 \u2013 x<sup>2<\/sup><\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Let u = x<sup>2<\/sup> + 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; du\/dx = 2x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v= 1 \u2013 x<sup>2&nbsp; <\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;dv\/dx = &#8211; 2x<\/p>\n\n\n\n<p>dy =&nbsp; vdu-&nbsp; udv<\/p>\n\n\n\n<p>dxdxdx<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"72\" height=\"3\"><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v<sup>2<\/sup><\/strong><\/p>\n\n\n\n<p>dy\/dx = (1 \u2013 x<sup>2<\/sup>)(2x) \u2013 (x<sup>2<\/sup> + 1)(-2x)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (1 \u2013 x<sup>2<\/sup>)<sup>2<\/sup><\/p>\n\n\n\n<p>=&nbsp; 2x \u2013 2x<sup>3<\/sup> + 2x<sup>3<\/sup> + 2x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (1 \u2013 x<sup>2<\/sup>)<sup>2<\/sup><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"53\" height=\"2\"><strong>dy\/dx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 4x<\/strong><\/p>\n\n\n\n<ul><li>\u2013 x<sup>2<\/sup>)<sup>2<\/sup><\/li><\/ul>\n\n\n\n<ul><li>y = (x \u2013 1)<sup>2<\/sup><\/li><\/ul>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u221ax<\/p>\n\n\n\n<p>Let u = (x \u2013 1)<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; du\/dx = 2(x \u2013 1)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v = \u221ax&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; dv\/dx = 1\/2\u221ax<\/p>\n\n\n\n<p>dy\/dx = \u221ax 2(x &#8211; 1) -(x \u2013 1)<sup>2<\/sup> 1\/2\u221ax&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (\u221ax)<sup>2<\/sup><\/p>\n\n\n\n<p>dy\/dx = \u221ax 2(x &#8211; 1) &#8211; (x \u2013 1)<sup>2<\/sup> 1\/2\u221ax&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>x<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"57\" height=\"2\"><strong>Evaluation:&nbsp; <\/strong>Differentiate with respect to x: (1)&nbsp; (2x + 3)<sup>3&nbsp; <\/sup>&nbsp;&nbsp;(2)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u221ax<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (x<sup>3<\/sup>\u2013 4)<sup>2<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u221a(x + 1)<\/p>\n\n\n\n<p><strong>Applications of differentiation:<\/strong><\/p>\n\n\n\n<p>There are many applications of differential calculus.<\/p>\n\n\n\n<p>Examples:<\/p>\n\n\n\n<ol type=\"1\"><li>Find the gradient of the curve y = x<sup>3 <\/sup>\u2013 5x<sup>2 <\/sup>+ 6x \u2013 3 at the point where x = 3.<\/li><\/ol>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Solution:<\/p>\n\n\n\n<p>&nbsp;Y = x<sup>3 <\/sup>\u2013 5x<sup>2 <\/sup>+ 6x \u2013 3<\/p>\n\n\n\n<p>dy\/dx = 3x<sup>2<\/sup> \u2013 10x + 6<\/p>\n\n\n\n<p>where x = 3; dy\/dx = 3(3<sup>2<\/sup>) \u2013 10(3) + 6<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 27 \u2013 30 + 6<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 3.<\/p>\n\n\n\n<ul><li>Find the coordinates of the point on the graph of y = 5x<sup>2 <\/sup>+ 8x \u2013 1 at which the gradient is \u2013 2<\/li><\/ul>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp; Y = 5x<sup>2<\/sup> + 8x \u2013 1<\/p>\n\n\n\n<p>dy\/dx = 10x + 8<\/p>\n\n\n\n<p>replace; dy\/dx by \u2013 2<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10x + 8 = &#8211; 2<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10x = &#8211; 2 \u2013 8<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x = -10\/10&nbsp;&nbsp; = &#8211; 1<\/p>\n\n\n\n<ul><li>Find the point at which the tangent to the curve y = x<sup>2 <\/sup>&nbsp;&#8211; 4x + 1 at the point (2, -3)<\/li><\/ul>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Y =x<sup>2 <\/sup>&nbsp;&#8211; 4x + 1<\/p>\n\n\n\n<p>dy\/dx = 2x \u2013 4<\/p>\n\n\n\n<p>at point (2, -3): dy\/dx = 2(2) \u2013 4<\/p>\n\n\n\n<p>dy\/dx = 0<\/p>\n\n\n\n<p>tangent to the curve:&nbsp; y \u2013 y1 = dy\/dx(x \u2013 x1)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y \u2013 (-3) = 0 ( x- 2)<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;y + 3 = 0<\/p>\n\n\n\n<p><strong>Evaluation<\/strong>:&nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1. Find the coordinates of the point on the graph of y = x<sup>2<\/sup> + 2x \u2013 10 at which the gradient is 8.<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2. Find the point on the curve y = x<sup>3<\/sup> + 3x<sup>2<\/sup> \u2013 9x + 3 at which the gradient is 15.<\/p>\n\n\n\n<p><strong>Velocity and Acceleration<\/strong><\/p>\n\n\n\n<p><strong>Velocity<\/strong>: The velocity after t seconds is the rate of change of displacement with respect to time.<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Suppose; s = distance and t = time,<\/p>\n\n\n\n<p>Then;&nbsp; <strong><em>Velocity = ds\/dt<\/em><\/strong><\/p>\n\n\n\n<p><strong>Acceleration<\/strong>:&nbsp; This is the rate of change of velocity compared with time.<\/p>\n\n\n\n<p><strong><em>Acceleration = dv\/dt<\/em><\/strong><\/p>\n\n\n\n<p>&nbsp;Example:<\/p>\n\n\n\n<p>A moving body goes s metres in t seconds, where s = 4t<sup>2<\/sup> \u2013 3t + 5. Find its velocity after 4 seconds. Show that the acceleration is constant and find its value.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; S = 4t<sup>2<\/sup> \u2013 3t + 5<\/p>\n\n\n\n<p>ds\/dt = 8t \u2013 3<\/p>\n\n\n\n<p>velocity = ds\/dt = 8(4) \u2013 3<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 32 \u2013 3<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 29<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Acceleration: dv\/dt = 8.<\/p>\n\n\n\n<p><strong>Maxima and Minimal<\/strong><\/p>\n\n\n\n<ol type=\"1\"><li>Find the maximum and minimum value of y on the curve 6x \u2013 x<sup>2<\/sup>.<\/li><\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = 6x \u2013 x<sup>2<\/sup><\/p>\n\n\n\n<p>dy\/dx = 6 \u2013 2x<\/p>\n\n\n\n<p>equatedy\/dx = 0<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6 \u2013 2x = 0<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6 = 2x<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; X = 3<\/p>\n\n\n\n<p>The &nbsp;turning point is (3, 9)<\/p>\n\n\n\n<ul><li>Find the maximum and minimum of the function x<sup>3<\/sup> \u2013 12x + 2.<\/li><\/ul>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Y =x<sup>3<\/sup> \u2013 12x + 2<\/p>\n\n\n\n<p>dy\/dx = 3x<sup>2<\/sup> \u2013 12<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3x<sup>2<\/sup> \u2013 12 = 0<\/p>\n\n\n\n<p>&nbsp;3x<sup>2<\/sup> = 12<\/p>\n\n\n\n<p>x<sup>2<\/sup> = 12\/3<\/p>\n\n\n\n<p>x<sup>2<\/sup> = 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x = \u00b1 2<\/p>\n\n\n\n<p>minimum point occur when d<sup>2<\/sup>y\/dx<sup>2<\/sup>&gt; 0<\/p>\n\n\n\n<p>maximum point occurs when d<sup>2<\/sup>y\/dx<sup>2<\/sup>&lt; 0<\/p>\n\n\n\n<p>d<sup>2<\/sup>y\/dx<sup>2<\/sup>= 6x<\/p>\n\n\n\n<p>substitute x = 2;&nbsp;&nbsp; d<sup>2<\/sup>y\/dx<sup>2<\/sup> = 6 x 2 = 12<\/p>\n\n\n\n<p>therefore: the function is minimum at point&nbsp; x = 2 and y = &#8211; 14<\/p>\n\n\n\n<p>substitute x = &#8211; 2; d<sup>2<\/sup>y\/dx<sup>2<\/sup> = 6(-2) = -12<\/p>\n\n\n\n<p>therefore:&nbsp; the function is maximum at point x = &#8211; 2 and y = 18<\/p>\n\n\n\n<p><strong>Evaluation:<\/strong><\/p>\n\n\n\n<p>1. A particle moves in such a way that after t seconds it has gone s metres, where s = 5t + 15t<sup>2<\/sup> \u2013 t<sup>3<\/sup><\/p>\n\n\n\n<p>2. Find the maximum and minimum value of y on the curve 4 \u201312x &#8211; 3x<sup>2<\/sup>.<\/p>\n\n\n\n<p><strong>General Evaluation<\/strong><\/p>\n\n\n\n<p>Use product rule to find the derivative of<\/p>\n\n\n\n<p>1.&nbsp;&nbsp; y = x<sup>2<\/sup> (1 + x)<sup>\u00bd<\/sup><\/p>\n\n\n\n<p>2.&nbsp;&nbsp; y = \u221ax (x<sup>2<\/sup> + 3x \u2013 2)<sup>2<\/sup><\/p>\n\n\n\n<p>3.&nbsp;&nbsp; Find&nbsp; the&nbsp; derivative&nbsp;&nbsp; of&nbsp;&nbsp; y =(7x<sup>2<\/sup> -5)<sup>3<\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>4. Using completing the square method find t if s=ut+1at<sup>2<\/sup><\/p>\n\n\n\n<p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p>5. If 3 is a root of the equation x<sup>2 <\/sup>\u2013 kx +42=0 find the value of k and the other root of the equation<\/p>\n\n\n\n<p><strong>READING ASSIGNMENT: NGM for SS 3 Chapter 10 <\/strong>page 90 -101,<\/p>\n\n\n\n<p><strong>WEEKEND ASSIGNMENT<\/strong><\/p>\n\n\n\n<p><strong>OBJECTIVE<\/strong><\/p>\n\n\n\n<p>1.Differentiate the function&nbsp; 4x<sup>4<\/sup> + x<sup>3<\/sup> \u2013 5 (a)4x<sup>3<\/sup> +3x<sup>2<\/sup>&nbsp;&nbsp; (b)16x<sup>2<\/sup> +3x<sup>2<\/sup> (c)16x<sup>3<\/sup> +3x<sup>2<\/sup> (d)16x<sup>4<\/sup> +&nbsp; 3x<sup>2<\/sup><\/p>\n\n\n\n<p>2.Find d<sup>2<\/sup>y\/dx<sup>2<\/sup>&nbsp;&nbsp; of the function y = 3x<sup>5<\/sup>wrt x. (a) 15x<sup>3<\/sup> (b) 45 x<sup>4<\/sup> (c) 60x<sup>3<\/sup> (d) 3x<sup>5<\/sup> (e) 12x<sup>3<\/sup><\/p>\n\n\n\n<p>3.If f(x) = 3x<sup>2<\/sup> + 2\/x find f<sup>1<\/sup>(x)&nbsp;&nbsp; (a) 6x + 2&nbsp;&nbsp;&nbsp; (b) 6x + 2\/x<sup>2<\/sup>&nbsp;&nbsp;&nbsp; (c) 6x \u2013 2\/x<sup>2<\/sup>&nbsp; (d)6x -2<\/p>\n\n\n\n<p>4.Find the derivative of 2x<sup>3<\/sup> \u2013 6x<sup>2<\/sup>&nbsp;&nbsp;&nbsp; (a) 6x<sup>2<\/sup> \u2013 12x&nbsp;&nbsp;&nbsp; (b) 6x<sup>2<\/sup> \u2013 12x&nbsp; (c) 2x<sup>2<\/sup> \u2013 6x&nbsp; (d) 8x<sup>2<\/sup> \u2013 3x<\/p>\n\n\n\n<p>5.Find the derivative of x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> + 15x&nbsp;&nbsp; (a) x<sup>2<\/sup> \u2013 7x + 15 (b) 3x<sup>2<\/sup> \u2013 14x + 15 (c) 3x<sup>2<\/sup> + 7x + 15 (d) 3x<sup>2<\/sup> \u2013 7x + 15<\/p>\n\n\n\n<p><strong>THEORY<\/strong><\/p>\n\n\n\n<p>1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Differentiate with respect to x. y<sup>2<\/sup> + x<sup>2<\/sup> \u2013 3xy = 4<\/p>\n\n\n\n<p>2.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Find the derivative of 3x<sup>3<\/sup>(x<sup>2<\/sup> + 4)<sup>2<\/sup><\/p>\n\n\n\n<p><\/p>\n\n\n\n<p>Mathematics Topics for Senior Secondary School &#8211;<a href=\"https:\/\/enoteprime.com\" target=\"_blank\" rel=\"noreferrer noopener\"> Enoteprime.com<\/a><\/p>\n\n\n\n<p><strong>WEEK SIX<\/strong><strong><\/strong><\/p>\n\n\n\n<p>INTEGRATION OF ALGEBRAIC FUNCTION<\/p>\n\n\n\n<p>INTEGRATION: Is defined as the anti-differentiation or is the process from the derivative form to the anti-derivative form.<\/p>\n\n\n\n<p>Y = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"34\" height=\"30\">&nbsp;+ c, where a is the coefficient of x. n is the power of x and c is the constant value through differentiation.<\/p>\n\n\n\n<p>NOTE: when integrating <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"7\" height=\"27\">&nbsp;= ln x + c.<\/p>\n\n\n\n<p>EXAMPLE<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"15\" height=\"27\">&nbsp;= 3<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"17\" height=\"19\">&#8211; x + 5, find y?<\/p>\n\n\n\n<p>SOLUTION<\/p>\n\n\n\n<p>Y = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"34\" height=\"30\">&nbsp;+ c. <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"9\" height=\"19\">&nbsp; y = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"129\" height=\"30\">&nbsp;+ c<\/p>\n\n\n\n<p>Y = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"20\" height=\"30\">&nbsp;<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"31\" height=\"30\">&nbsp;7+ <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"14\" height=\"27\">&nbsp;+ c = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"49\" height=\"30\">&nbsp;+ 5x + c.<\/p>\n\n\n\n<p>STANDARD INTEGRAL FORMS<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"62\" height=\"20\">&nbsp;= G(x) + c<\/p>\n\n\n\n<p>EXAMPLE<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"\" alt=\"\"\/><\/figure>\n\n\n\n<p>SOLUTION<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"51\" height=\"26\">&nbsp;= <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"27\" height=\"44\">&nbsp;+ C<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp; <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"18\" height=\"44\">&nbsp;+ C&nbsp; <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"9\" height=\"19\">&nbsp;<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"21\" height=\"37\">&nbsp;+ C<\/p>\n\n\n\n<p>INTEGRATION BY ALGEBRAIC SUBSTITUTION<\/p>\n\n\n\n<p>EXAMPLE<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"\" alt=\"\"\/><\/figure>\n\n\n\n<p>SOLUTION<\/p>\n\n\n\n<p>Let u = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"17\" height=\"19\">&nbsp;, <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"15\" height=\"27\">&nbsp;= <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"25\" height=\"19\">&nbsp;<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"9\" height=\"19\">&nbsp;<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"18\" height=\"19\">&nbsp;= <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"20\" height=\"28\"><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"46\" height=\"20\">.<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"20\" height=\"28\">&nbsp;= <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"56\" height=\"27\"><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"59\" height=\"27\"><em>&nbsp;=<\/em> <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"26\" height=\"27\">+ c = <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"19\" height=\"33\">&nbsp;+ c.<\/p>\n\n\n\n<p>AREA UNDER THE CURVE<\/p>\n\n\n\n<p>The integral <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"68\" height=\"26\">&nbsp; is called the definite integral of the function f(x) with a and b the lower and upper limits of the integral.<\/p>\n\n\n\n<p>EXAMPLE<\/p>\n\n\n\n<p>Evaluate<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"62\" height=\"26\">.<\/p>\n\n\n\n<p>SOLUTION<\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"44\" height=\"26\">&nbsp;=[ <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"32\" height=\"30\">&nbsp;+ c<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"13\" height=\"19\"><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"148\" height=\"29\">&nbsp;= <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"130\" height=\"29\"><\/p>\n\n\n\n<p><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"58\" height=\"27\">&nbsp;= <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"26\" height=\"27\">&nbsp;= 913.5<\/p>\n\n\n\n<p>ASSESSMENT: Integrate the following:<\/p>\n\n\n\n<ol type=\"1\"><li><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"100\" height=\"26\"><\/li><li><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"79\" height=\"30\">)<img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"18\" height=\"19\"><\/li><li><img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"59\" height=\"22\"><\/li><li>&nbsp;Using substitution method, evaluate <img decoding=\"async\" loading=\"lazy\" src=\"\" width=\"60\" height=\"30\">.<\/li><\/ol>\n\n\n\n<p>Mathematics Topics for Senior Secondary School &#8211;<a href=\"https:\/\/enoteprime.com\" target=\"_blank\" rel=\"noreferrer noopener\"> Enoteprime.com<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculation on interest on bonds and debentures, Coordinate Geometry of straight line, Mathematics Topics for Senior Secondary School<\/p>\n","protected":false},"author":1,"featured_media":13512,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[71,4819,4821,4854,4858,4856,4860],"tags":[2739,209,2587,4898,4900],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v20.4 (Yoast SEO v20.4) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Enote on Mathematics SS3 Second Term - Enoteprime.com<\/title>\n<meta name=\"description\" content=\"Calculation on interest on bonds and debentures, Coordinate Geometry of straight line, Mathematics Topics for Senior Secondary School\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/enoteprime.com\/Enote-on-mathematics-ss3-second-term\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta 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