{"id":4083,"date":"2023-10-06T09:29:23","date_gmt":"2023-10-06T09:29:23","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=4083"},"modified":"2023-10-06T09:30:04","modified_gmt":"2023-10-06T09:30:04","slug":"week-5-ss3-second-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-ss3-second-term-mathematics-notes\/","title":{"rendered":"Week 5 &#8211; SS3 Second Term Mathematics Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK FIVE<br \/>\n\t\t\t<\/strong><\/p>\n<ul>\n<li>Differentiation of algebraic functions:\n<\/li>\n<li>Basic rules of differentiation such as sum and difference, product rule, quotient rule\n<\/li>\n<li>Maximal and minimum application.\n<\/li>\n<\/ul>\n<p>\u00a0<strong>Derivative of algebraic functions<br \/>\n<\/strong>Let f, u, v be functions such that<br \/>\nf(x) = u(x) + v(x)<br \/>\nf(x +x ) = u(x +x ) + v(x + x)<br \/>\nf(x + x) \u2013 f(x) = {u(x+ x) + v(x+ x) \u2013 v(x + x) \u2013 u(x) \u2013 v(x)}<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= u (x +x ) \u2013 u(x) + v(x +x ) \u2013 v(x)<br \/>\nf(x + x) \u2013 f(x) = u(x +x)-u(x) + v(x +x ) \u2013 v(x)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se1.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se2.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se3.png\" alt=\"\"\/><br \/>\n\t\tLim  f(x + x) \u2013 f(x) = U<sup>1<\/sup>(x) + V<sup>1<\/sup>(x)<br \/>\nif y = u + v and u and v are functions of x, then <strong>dy\/dx = du\/dx + dv\/dx<\/strong><\/p>\n<p>\u00a0<strong>Examples<\/strong>:Find the derivative of the following<br \/>\n1) 2x<sup>3<\/sup> \u2013 5 x<sup>2<\/sup> + 2          2)3x<sup>2<\/sup> + 1\/x             3)2x<sup>3<\/sup> + 2x<sup>2<\/sup> +1<\/p>\n<p>\u00a0Solution<br \/>\n1.\u00a0\u00a0\u00a0\u00a0Let y = 2x<sup>3<\/sup> \u2013 5x<sup>2<\/sup> + 2<br \/>\n\u00a0\u00a0\u00a0\u00a0dy\/dx = 6x<sup>2<\/sup> \u2013 10x<\/p>\n<p>\u00a02.\u00a0\u00a0\u00a0\u00a0Let y = 3x<sup>2<\/sup>+ 1\/x = 3x<sup>2<\/sup> + x<sup>-1<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0dy\/dx = 6x \u2013 x<sup>-2<\/sup> = 6x \u2013 1<br \/>\n\t\tx<sup>2<\/sup><br \/>\n\t\t3.        Let y = 2x<sup>3<\/sup> + 2x<sup>2<\/sup> + 1<br \/>\ndy\/dx=6x<sup>2<\/sup>  +  4x<strong><br \/>\n\t\t\t<\/strong><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong><br \/>\n\t\t<strong>Evaluation:   <\/strong>1. If y = 3x<sup>4<\/sup> \u2013 2x<sup>3<\/sup> \u2013 7x + 5. Find dy\/dx<strong><br \/>\n\t\t\t<\/strong>2.Findd (8x<sup>3<\/sup> \u2013 5x<sup>2<\/sup> + 6)<br \/>\nDx<\/p>\n<p>\u00a0<strong>Function of a function (chain rule)<br \/>\n<\/strong>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?<br \/>\nGiven that y = f(x) and u = h(x), what is dy\/dx?<br \/>\ndy\/dx = ,   this is called the chain rule<\/p>\n<p>\u00a0<strong>Examples<\/strong><br \/>\n\t\tFind the derivative of the following.(a)y = (3x<sup>2<\/sup> \u2013 2)<sup>3<\/sup>   (b)  y = (1 \u2013 2x<sup>3<\/sup>)  (c) 5\/(6-x<sup>2<\/sup>)<sup>3<\/sup><br \/>\n\t\tSolution<br \/>\n1.\u00a0\u00a0\u00a0\u00a0y = (3x<sup>2<\/sup> \u2013 2)<sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0Let u = 3x<sup>2<\/sup> \u2013 2<br \/>\n\u00a0\u00a0\u00a0\u00a0y = (3x<sup>2<\/sup> \u2013 2)<sup>3<\/sup> =&gt; y = u<sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y = u<sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0dy\/du = 3u<sup>2<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0du\/dx = 6x<br \/>\n\u00a0\u00a0\u00a0\u00a0<sup><br \/>\n\t\t\t<\/sup>\u00a0\u00a0\u00a0\u00a0dy\/dx =  = 3u<sup>2<\/sup> x 6x<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 18xu<sup>2<\/sup> = 18x(3x<sup>2<\/sup> \u2013 2)<sup>2<\/sup><\/p>\n<p>\u00a02.\u00a0\u00a0\u00a0\u00a0y = (1 \u2013 2x<sup>3<\/sup>)<sup>1\/2<\/sup> =&gt; (1 \u2013 2x<sup>3<\/sup>) <sup>1\/2<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0Let u =1 \u2013 2x<sup>3<\/sup>,     hence  y = u<sup>1\/2 <\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0dy\/dx = = \u00bd u<sup>-1\/2<\/sup>  x( \u20136x<sup>2<\/sup>)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= -3x<sup>2<\/sup> u <sup>\u2013 \u00bd<\/sup>=  -3x<sup>2<\/sup><br \/>\n\t\tu<sup>1\/2<\/sup><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se4.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se5.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0     -3x<sup>2<\/sup>\u00a0\u00a0\u00a0\u00a0=        -3x<sup>2<\/sup>\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se6.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se7.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    \u221a u\u00a0\u00a0\u00a0\u00a0   \u221a(1 \u2013 2x<sup>3<\/sup>)<\/p>\n<p>\u00a03.\u00a0\u00a0\u00a0\u00a0y =      5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 5(6 \u2013 x<sup>2<\/sup>)<sup>-3<\/sup><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se8.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0        (6 \u2013 x<sup>2<\/sup>)<sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0Let u = 6 \u2013 x<sup>2<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0y = 5(u)<sup>\u20133<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0dy\/du = -15u <sup>\u20134<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0du\/dx = -2x<br \/>\n\u00a0\u00a0\u00a0\u00a0dy\/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>\u00a0\u00a0\u00a0\u00a0<br \/>\n            =          30x_<br \/>\n                     (6 \u2013 x<sup>2<\/sup>)<sup>4<\/sup><\/p>\n<p>\u00a0<strong>Evaluation<\/strong>:<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se9.png\" alt=\"\"\/>1.\u00a0\u00a0\u00a0\u00a0Given that y =       1                find dy\/dx<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0              (2x + 3)<sup>4<\/sup><br \/>\n\t\t2.\u00a0\u00a0\u00a0\u00a0If y = (3x<sup>2<\/sup> + 1)<sup>3<\/sup> , Find dy\/dx<\/p>\n<p>\u00a0<strong>Product Rule<br \/>\n<\/strong>We shall consider the derivative of y = uv where u and v are function of x.<strong><br \/>\n\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0Let y = uv<br \/>\nThen y + y = (u +u )(v + v)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= uv + uv + vu +uv<br \/>\n\u00a0\u00a0\u00a0\u00a0y = uv + uv + vu+ uv \u2013 uv<br \/>\n\u00a0\u00a0\u00a0\u00a0y= uv + vu + uv<br \/>\n\u00a0\u00a0\u00a0\u00a0y=  uv  +    vu   +    uv<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x\u00a0\u00a0\u00a0\u00a0x\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0As x =&gt;0 ,u=&gt; 0 , v=&gt; 0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se10.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0          Lim       y   =     Lim       uv +    Lim        vu   +     Lim        uv<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se11.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se12.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se13.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0x=&gt;0   x          x=&gt;0    x        x=&gt;0      x           x=&gt;0      x<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0Hence   <strong>dy\/dx<\/strong>= <strong>U dv + Vdu<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se14.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se15.png\" alt=\"\"\/><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0dxdx<br \/>\n<\/strong><br \/>\n\u00a0<strong>Examples<br \/>\n<\/strong>Find the derivatives of the following.<br \/>\n(a)\u00a0\u00a0\u00a0\u00a0y = (3 + 2x) (1 \u2013 x)               (b) y = (1 \u2013 2x + 3x<sup>2<\/sup>) (4 \u2013 5x<sup>2<\/sup>)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0Solution<br \/>\n1.\u00a0\u00a0\u00a0\u00a0y = (3 + 2x) (1 \u2013 x)<br \/>\n\u00a0\u00a0\u00a0\u00a0Let u = 3 + 2x and v = (1 \u2013 x)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0du\/dx = 2 and dv\/dx = -1<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0dv\/dx = u dv + vdu<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se16.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0dx\u00a0\u00a0\u00a0\u00a0dx<br \/>\n=  (1 \u2013 x) 2 + (3 + 2x) (-1)     = 2 \u2013 2x \u2013 3 \u2013 2x<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0dy\/dx\u00a0\u00a0\u00a0\u00a0= &#8211; 1 \u2013 4x<\/p>\n<p>\u00a02.\u00a0\u00a0\u00a0\u00a0y = (1 \u2013 2x + 3x<sup>2<\/sup>) (4 \u2013 5x<sup>2<\/sup>)<br \/>\n\u00a0\u00a0\u00a0\u00a0Let u = (1 \u2013 2x + 3x<sup>2<\/sup>)         and v = (4 \u2013 5x<sup>2<\/sup>)<br \/>\n\u00a0\u00a0\u00a0\u00a0du\/dx = -2 + 6x                      and dv\/dx = &#8211; 10x<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se17.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0dy\/dx = udv +   vdu<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se18.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0dxdx<br \/>\n\u00a0\u00a0\u00a0\u00a0= (1 \u2013 2x + 3x<sup>2<\/sup>) (-10x) + (4 \u2013 5x<sup>2<\/sup>) (- 2 + 6x)<br \/>\n\u00a0\u00a0\u00a0\u00a0= &#8211; 10x + 20x<sup>2<\/sup> \u2013 30x<sup>3<\/sup> + (- 8 + 10x<sup>2<\/sup> + 24x \u2013 30x<sup>3<\/sup>)<br \/>\n\u00a0\u00a0\u00a0\u00a0= &#8211; 10x + 20x<sup>2<\/sup> \u2013 30x<sup>3<\/sup> \u2013 8 + 10x<sup>2<\/sup> + 24x \u2013 30x<sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0= 14x + 30x<sup>2<\/sup> \u2013 60x<sup>3<\/sup> \u2013 8<br \/>\n<sup>\u00a0\u00a0\u00a0\u00a0<\/sup><br \/>\n\t\t<strong>Evaluation<br \/>\n<\/strong>Given that (i) y = (5+3x)(2-x)    (ii)  y = (1+x)(x+2)<sup>3\/2<\/sup><strong> ,<\/strong>find dy\/dx<\/p>\n<p>\u00a0<strong>Quotient Rule:<br \/>\n<\/strong>If y = u<br \/>\n<strong>v<\/strong><br \/>\n\t\tthen; <strong>dy =  vdu- udv<br \/>\n\t\t\t\t<\/strong><strong>dxdxdx<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se19.png\" alt=\"\"\/><strong>                          v<sup>2<\/sup><br \/>\n\t\t\t<\/strong><strong>E<\/strong>xamples:<br \/>\nDifferentiate the following with respect to x.  (a)x<sup>2<\/sup> + 1     (b)(x \u2013 1)<sup>2<\/sup><br \/>\n\t\t\t                                                                                              1 \u2013 x<sup>2<\/sup>               \u221ax<br \/>\nSolution:<\/p>\n<ol>\n<li>y = x<sup>2<\/sup> + 1\n<\/li>\n<\/ol>\n<p>                      1 \u2013 x<sup>2<\/sup><br \/>\n\t\t             Let u = x<sup>2<\/sup> + 1      du\/dx = 2x<br \/>\n                   v= 1 \u2013 x<sup>2  <\/sup>        dv\/dx = &#8211; 2x<\/p>\n<p>\u00a0dy =  vdu-  udv<br \/>\n\t\tdxdxdx<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se20.png\" alt=\"\"\/><strong>                v<sup>2<\/sup><br \/>\n\t\t\t<\/strong>dy\/dx = (1 \u2013 x<sup>2<\/sup>)(2x) \u2013 (x<sup>2<\/sup> + 1)(-2x)<br \/>\n                                             (1 \u2013 x<sup>2<\/sup>)<sup>2<br \/>\n<\/sup>=  2x \u2013 2x<sup>3<\/sup> + 2x<sup>3<\/sup> + 2x<br \/>\n\t\t                 (1 \u2013 x<sup>2<\/sup>)<sup>2<br \/>\n<\/sup><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se21.png\" alt=\"\"\/><strong>dy\/dx        =                        4x<br \/>\n<\/strong><\/p>\n<ol>\n<li>\u2013 x<sup>2<\/sup>)<sup>2<br \/>\n<\/sup><\/li>\n<\/ol>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>y = (x \u2013 1)<sup>2<\/sup>\n\t\t\t\t\t<\/div>\n<p>          \u221ax<br \/>\nLet u = (x \u2013 1)<sup>2<\/sup>        du\/dx = 2(x \u2013 1)<br \/>\n        v = \u221ax                    dv\/dx = 1\/2\u221ax<br \/>\ndy\/dx = \u221ax 2(x &#8211; 1) -(x \u2013 1)<sup>2<\/sup> 1\/2\u221ax<br \/>\n                                  (\u221ax)<sup>2<\/sup><br \/>\n\t\t\t\tdy\/dx = \u221ax 2(x &#8211; 1) &#8211; (x \u2013 1)<sup>2<\/sup> 1\/2\u221ax<br \/>\nx<\/p>\n<p>\u00a0<\/li>\n<\/ol>\n<p><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0929_Week5SS3Se22.png\" alt=\"\"\/><strong>Evaluation:  <\/strong>Differentiate with respect to x: (1)  (2x + 3)<sup>3  <\/sup>  (2)         \u221ax<br \/>\n\t\t\t                                                                                (x<sup>3<\/sup>\u2013 4)<sup>2<\/sup>                 \u221a(x + 1)<\/p>\n<p>\u00a0<strong>Applications of differentiation:<br \/>\n<\/strong>There are many applications of differential calculus.<\/p>\n<p>\u00a0Examples:<\/p>\n<ol>\n<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.\n<\/li>\n<\/ol>\n<p>      Solution:<br \/>\n Y = x<sup>3 <\/sup>\u2013 5x<sup>2 <\/sup>+ 6x \u2013 3<br \/>\ndy\/dx = 3x<sup>2<\/sup> \u2013 10x + 6<br \/>\nwhere x = 3; dy\/dx = 3(3<sup>2<\/sup>) \u2013 10(3) + 6<br \/>\n                                      = 27 \u2013 30 + 6<br \/>\n                                      = 3.<\/p>\n<ol>\n<li>\n<div>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\n<\/div>\n<p>Solution:<br \/>\n   Y = 5x<sup>2<\/sup> + 8x \u2013 1<br \/>\ndy\/dx = 10x + 8\n<\/li>\n<\/ol>\n<p>replace; dy\/dx by \u2013 2<br \/>\n                              10x + 8 = &#8211; 2<br \/>\n                               10x = &#8211; 2 \u2013 8<br \/>\n                           x = -10\/10   = &#8211; 1 <\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>Find the point at which the tangent to the curve y = x<sup>2 <\/sup> &#8211; 4x + 1 at the point (2, -3)\n<\/div>\n<p>Solution:<br \/>\n                 Y =x<sup>2 <\/sup> &#8211; 4x + 1\n<\/li>\n<\/ol>\n<p>dy\/dx = 2x \u2013 4<br \/>\nat point (2, -3): dy\/dx = 2(2) \u2013 4<br \/>\ndy\/dx = 0<br \/>\ntangent to the curve:  y \u2013 y1 = dy\/dx(x \u2013 x1)<br \/>\n                                                                  y \u2013 (-3) = 0 ( x- 2)<br \/>\n                                                                  y + 3 = 0<\/p>\n<p>\u00a0<strong>Evaluation<\/strong>:<br \/>\n               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.<br \/>\n               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<p>\u00a0<br \/>\n\u00a0<strong>Velocity and Acceleration<br \/>\n<\/strong><strong>Velocity<\/strong>: The velocity after t seconds is the rate of change of displacement with respect to time.<br \/>\n               Suppose; s = distance and t = time,<br \/>\nThen;  <strong><em>Velocity = ds\/dt<\/em><\/strong><\/p>\n<p>\u00a0<strong>Acceleration<\/strong>:  This is the rate of change of velocity compared with time.<br \/>\n<strong><em>Acceleration = dv\/dt<br \/>\n<\/em><\/strong> Example:<br \/>\nA 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.<br \/>\nSolution:<br \/>\n                     S = 4t<sup>2<\/sup> \u2013 3t + 5<br \/>\nds\/dt = 8t \u2013 3<br \/>\nvelocity = ds\/dt = 8(4) \u2013 3<br \/>\n                                                      = 32 \u2013 3<br \/>\n                                                     = 29<br \/>\n                      Acceleration: dv\/dt = 8.<\/p>\n<p>\u00a0<strong>Maxima and Minimal<br \/>\n<\/strong><\/p>\n<ol>\n<li>\n<div>Find the maximum and minimum value of y on the curve 6x \u2013 x<sup>2<\/sup>.\n<\/div>\n<p>Solution:\n<\/li>\n<\/ol>\n<p>                     y = 6x \u2013 x<sup>2<br \/>\n<\/sup>dy\/dx = 6 \u2013 2x<br \/>\nequatedy\/dx = 0<br \/>\n                     6 \u2013 2x = 0<br \/>\n                            6 = 2x<br \/>\n                            X = 3<br \/>\nThe  turning point is (3, 9)<\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>Find the maximum and minimum of the function x<sup>3<\/sup> \u2013 12x + 2.\n<\/div>\n<p>Solution:<br \/>\n                      Y =x<sup>3<\/sup> \u2013 12x + 2<br \/>\ndy\/dx = 3x<sup>2<\/sup> \u2013 12<br \/>\n                           3x<sup>2<\/sup> \u2013 12 = 0<br \/>\n 3x<sup>2<\/sup> = 12<br \/>\nx<sup>2<\/sup> = 12\/3<br \/>\nx<sup>2<\/sup> = 4                  x = \u00b1 2<br \/>\nminimum point occur when d<sup>2<\/sup>y\/dx<sup>2<\/sup>&gt; 0<br \/>\nmaximum point occurs when d<sup>2<\/sup>y\/dx<sup>2<\/sup>&lt; 0<br \/>\nd<sup>2<\/sup>y\/dx<sup>2<\/sup>= 6x<br \/>\nsubstitute x = 2;   d<sup>2<\/sup>y\/dx<sup>2<\/sup> = 6 x 2 = 12\n<\/li>\n<\/ol>\n<p>therefore: the function is minimum at point  x = 2 and y = &#8211; 14<br \/>\nsubstitute x = &#8211; 2; d<sup>2<\/sup>y\/dx<sup>2<\/sup> = 6(-2) = -12<br \/>\ntherefore:  the function is maximum at point x = &#8211; 2 and y = 18<\/p>\n<p>\u00a0<strong>Evaluation:<br \/>\n<\/strong>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<br \/>\n<\/sup>2. Find the maximum and minimum value of y on the curve 4 \u201312x &#8211; 3x<sup>2<\/sup>.<\/p>\n<p>\u00a0<strong>General Evaluation<\/strong><br \/>\n\t\tUse product rule to find the derivative of<br \/>\n1.   y = x<sup>2<\/sup> (1 + x)<sup>\u00bd <\/sup><br \/>\n\t\t2.   y = \u221ax (x<sup>2<\/sup> + 3x \u2013 2)<sup>2<\/sup><br \/>\n\t\t3.   Find  the  derivative   of   y =(7x<sup>2<\/sup> -5)<sup>3<\/sup>\u00a0\u00a0\u00a0\u00a0<br \/>\n4. Using completing the square method find t if s=ut+1at<sup>2<br \/>\n<\/sup><sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup>                                                 2\u00a0\u00a0\u00a0\u00a0<br \/>\n5. 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<p>\u00a0<strong>READING ASSIGNMENT: NGM for SS 3 Chapter 10 <\/strong>page 90 -101, <strong><br \/>\n\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong><strong>OBJECTIVE<br \/>\n<\/strong>1.Differentiate the function  4x<sup>4<\/sup> + x<sup>3<\/sup> \u2013 5 (a)4x<sup>3<\/sup> +3x<sup>2<\/sup>   (b)16x<sup>2<\/sup> +3x<sup>2<\/sup> (c)16x<sup>3<\/sup> +3x<sup>2<\/sup> (d)16x<sup>4<\/sup> +  3x<sup>2<\/sup><br \/>\n\t\t2.Find d<sup>2<\/sup>y\/dx<sup>2<\/sup>   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><br \/>\n\t\t3.If f(x) = 3x<sup>2<\/sup> + 2\/x find f<sup>1<\/sup>(x)   (a) 6x + 2    (b) 6x + 2\/x<sup>2<\/sup>    (c) 6x \u2013 2\/x<sup>2<\/sup>  (d)6x -2<br \/>\n4.Find the derivative of 2x<sup>3<\/sup> \u2013 6x<sup>2<\/sup>    (a) 6x<sup>2<\/sup> \u2013 12x    (b) 6x<sup>2<\/sup> \u2013 12x  (c) 2x<sup>2<\/sup> \u2013 6x  (d) 8x<sup>2<\/sup> \u2013 3x<br \/>\n5.Find the derivative of x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> + 15x   (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<p>\u00a0<strong>THEORY<br \/>\n\t\t\t\t<\/strong>1.\u00a0\u00a0\u00a0\u00a0Differentiate with respect to x. y<sup>2<\/sup> + x<sup>2<\/sup> \u2013 3xy = 4<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Find the derivative of 3x<sup>3<\/sup>(x<sup>2<\/sup> + 4)<sup>2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK FIVE Differentiation of algebraic functions: Basic rules of differentiation such as sum and difference,&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,316],"tags":[],"class_list":["post-4083","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4083","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=4083"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4083\/revisions"}],"predecessor-version":[{"id":4084,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4083\/revisions\/4084"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=4083"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=4083"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=4083"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}