kpk-fsc2-ch02-km.pdf

Exercise 2.1 Chapter 2  average rate of change h  ...

Exercise 2.1 Chapter 2  average rate of change h   h t  t   h  t  Chapter 2 With t  tmax  tmin t t then t  8.5  8  0.5 f  x  x   f  x  Substituting h  t   2t  7 Average rate of change y  x x h h  8.5  h  8  Exercise 2.1 t 0.5 Q1. Find the average rate of change of the h     2  8.5   7  2 8  7  following functions over the indicated intervals: t 0.5 a). y  x2  4 from x=2 to x=3 h 17  16  Sol: We have y  f  x   x  4 2 0.246 t 0.5 y f  x  x   f  x  Q2. Use definition to find out average rate of change  average rate of change  over specified interval for following functions: x x a). s  2t  3 from t=2 to t=5 With x  xmax  xmin then x  3  2  1 Solution: We have s  t   2t  3 y f  3  f  2   average rate of change s    s t  t   s  t  x x t t y 3  4    2  4  With t  tmax  tmin then t  5  2  3 2 2  x 1 s s  5   s  2   y t   9  4    4  4   13  8 3 x s  2  5   3   2  2   3  y t 3 5 x s 10  3  4  3  1 t 3 b). y  x 2  x from x=-3 to x=3 s 6 3  2 1 t 3 Solution: We have y  f  x   x 2  x s 3 2 t With x  xmax  xmin then x  3   3  6 b). y  x2  6x  8 from x=3 to x=3.1 y f  3  f  3  average rate of change  Solution: We have y  f  x   x  6 x  8 2 x 6  average rate of change y   f x  x   f  x   2 1   1   3  3  3    3  3  3  2 y x x  With x  xmax  xmin then x  3.1  3  0.1 x 6 y  9  1   9  1 y f  3.1  f  3   x 6 x 0.1 y 10  8 2  3.12  6  3.1  8   32  6  3  8   y     x 6 6  x 0.1 y y 9.61  18.6  8  9  18  8 1  x 3  x 0.1 c). s  2t 3  5t  7 from t=1 to t=3 y  0.99   1  Solution: We have s  t   2t  5t  7 3 x 0.1 y 0.99  1 0.01 1 With t  tmax  tmin then x  3  1  2     0.1 x 0.1 0.1 10 s s  3  s 1 A   r2  average rate of change  c). from r=2 to r=2.1 t 2 Solution: We have A  r    r 2 s  2  3  5  3  7    2 1  5 1  7   3   3   A A  r  r   A  r  t 2  average rate of change  r r s  2  27   15  7    2  5  7  t  2 With r  rmax  rmin then r  2.1  2  0.1 s 54  8  4 42 A A  2.1  A  2     t 2 2 r 0.1 s A   2.1    2 2 2  21  t r 0.1 d). h  2t  7 from t=8 to t=8.5 A 4.41  4  Solution: We have h  t   2t  7 r 0.1 Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 17  Exercise 2.2 Chapter 2 A 0.41 With t  tmax  tmin then t  5  3  2  r 0.1 p p  5   p  3 A  4.1 12.88  r t 2 p 3  5    5   1  3  3   3  1 d). h  t 9 from t=9 to t=16  2   2   Solution: We have h  t   t  9 t 2 With t  tmax  tmin then t  16  9  7 p 3  25   6   3  9   4    average rate of change h h 16   h  9  t 2  t 7 p 75  6  27  4         h 16 9 9 9 t      2 t 7 p 50 h 4  9  3  9   25  t 2 t 7 First principle for derivative h 4  3 1   dy y f  x  x   f  x  t 7 7  lim  lim Q3. A ball is thrown straight up. Its height after t seconds dx x0 x x0 x is given by the formula h  t   16t 2  80t Find the Slope when two point P1  x1 , y1  , P2  x2 , y2  are h given Slope = m  y2  y1 average velocity for the specified intervals x2  x1 t a). From t=2 to t=2.1 Differentiation of y  x n by first principle Rule Solution: We have h  t   16t  80t If f  x   x so f  x  x    x  x  2 n n With t  tmax  tmin then t  2.1  2  0.1 Using first principle for derivative of f  x  h h  2.1  h  2  f  x  x   f  x   average rate of change  d f  x   lim t 0.1 dx x 0 x h  16  2.1  80  2.1    16  2   80  2        x  x  2 2  xn n  f   x   lim t 0.1 x 0 x h 16  4.41  168  16  4   160 Using binomial theorem  n n 1 t  x  x   x n  nx n 1x  x n  2  x   n 2 0.1 2! h 70.56  168  64  160   x  x  n  xn t 0.1 so f   x   lim h 1.44 x 0 x   14.4 x  nx x   2!  x n2  x   n n 1 n 1  xn 2 t 0.1 n f  x   lim  b). From t=2 to t=2.01 x 0 x Solution: We have h  t   16t  80t 2 x n2  x   n n 1 nx n1x  2 With t  tmax  tmin then t  2.01  2  0.01 f   x   lim 2! x 0 x h h  2.1  h  2  x  nx  2! x  x    n 1 n n 1 n  2  average rate of change    t 0.1 f  x   lim  x 0 x  16  2.012  80  2.01    16  2 2  80  2   h     n n 1 n  2  f   x   lim  nx  2! x  x    n  1 t 0.01 x 0   h 16  4.0401  160.8  16  4   160 f   x   nx n1   2!  x n2  0   0  0   n n 1  t 0.01   h 64.6416  160.8  64  160  f   x   nx n 1 t 0.01 Exercise 2.2 h 0.1584   15.84 Q1. Use first Principle to determine the derivative t 0.01 of the following functions. Q4. The rate of change of price is called inflation. a). f  x   3x price P in rupees after t years is p  t   3t  t  1 2 Solution: We have f  x   3 x Find the average rate of change of inflation from t=3 to t=5 years. What the rate of change means? Explain First principle for derivative Solution: We have p  t   3t  t  1 2 dy y f  x  x   f  x   lim  lim p p  t  t   p  t  dx x0 x x0 x  average rate of change  Substituting f  x   3 x t t Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 18  Exercise 2.2 Chapter 2 dy y 3  x  x   3x y 12   x  x 2   12  x 2   lim  lim dy  lim  lim     dx x 0 x x 0 x dx x 0 x x 0 x dy  lim y  lim 3x  3x  3x dy  lim y  lim  12  x 2   x   2 xx  12  x 2 2  dx x 0 x x 0 x dx x 0 x x0 x y 3x 12  x 2   x   2 xx  12  x 2 dy 2  lim  lim dy y dx x 0 x x 0 x  lim  lim dy y dx x0 x x0 x  lim  lim 3  3   x   2 xx 2 dx  x x0 dy y  lim  lim x 0 b). f  x   5x  6 dx x0 x x0 x Solution: We have dy y x  x  2 x   lim  lim First principle for derivative dx x0 x x0 x y f  x  x   f  x  dy y  lim  x  2 x  dy  lim  lim  lim dx x 0 x x 0 x dx x0 x x0 Substituting dy y  lim  0  2 x y 5  x  x   6   5 x  6 x  lim  dy dx x 0  lim dx x 0 x x 0 x dy y y 5 x  5x  6  5 x  6  lim  2 x dy  lim  lim dx x 0 x dx x 0 x x f  x   16 x 2  7 x x 0 e). dy y 5x  lim  lim Solution: We have f  x   16 x  7 x 2 dx x 0 x x0 x dy y First principle for derivative  lim  lim 5 f  x  x   f  x  dx  x 0 x x0 dy y  lim  lim dy y dx x0 x x0 x  lim 5 Substituting dx x 0 x 16  x  x 2  7  x  x    16 x 2  7 x  y    c). f  x   x2  1 dy  lim  lim  dx x0 x x0 x Solution: We have f  x   x  1   2 16 x 2   x   2 xx  7 x  7x  16 x 2  7 x 2 dy y  lim  lim First principle for derivative dx x0 x x0 x dy y f  x  x   f  x  dy y 16 x 2  16  x   32 xx  7 x  7x  16 x 2  7 x 2  lim  lim  lim  lim dx x0 x x0 x dx x0 x x0 x 16  x   32 xx  7x 2 Substituting dy y  lim  lim  x  x 2  1   x 2  1 dx x 0 x x 0 x y     lim  dy y x 16x  32 x  7   lim dy  lim  lim dx x 0 x x 0 x dx x0 x x0 x y x 2   x   2 xx  1  x 2  1 2 dy y dy  lim  lim  lim  lim 16x  32 x  7  dx x 0 x x 0 x dx x0 x x0 dy y  x   2 xx  16  0   32 x  7  2 y  lim dx x0 x  dy  lim  lim dx x 0 x x 0 x dy y dy y x  x  2 x   lim  32 x  7  lim  lim dx x0 x dx x 0 x x 0 x 7 dy y f). f  x   lim  lim  x  2 x  x dx x 0 x x 0 7 dy y Solution: We have f  x    lim  0  2 x x dx x 0 x First principle for derivative dy y y f  x  x   f  x   lim  2x dy  lim  lim dx x 0 x dx x0 x x0 x d). f  x   12  x 2 7 Substituting f  x   Solution: We have f  x   12  x 2 x dy y 1  7 7 First principle for derivative  lim  lim    dy y f  x  x   f  x  dx x 0 x x 0 x  x  x x   lim  lim dx x0 x x0 x dy y 1  7 x  7  x  x    lim  lim   Substituting f  x   12  x x x  x  x  x   2   dx x 0 x 0 Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 19  Exercise 2.2 Chapter 2 dy y 1  7 x  7 x  7x  dy y 1  5  2 x  4   5  2 x  2x  4    lim  lim  lim  lim     dx x 0 x x0 x   2 x  2x  4  2 x  4   dx x0 x x0 x  x  x  x   y 1 10 x  20  10 x  10x  20  1  7x  dy dy y  lim  lim    lim  lim   dx x 0 x x0 x   2 x  2x  4  2 x  4   dx x0 x x0 x  x  x  x   dy y 1  10x  y  7   lim  lim   dx x0 x x 0 x   2 x  2x  4  2 x  4   dy  lim  lim   dx x0 x x0  x  x  x   dy y  10  y  7   7   lim  lim   dx x0 x x 0   2 x  2x  4  2 x  4   dy  lim    dx x0 x  x  x  0    x  x   dy y  10  dy y 7  lim    lim  dx x0 x   2 x  2  0   4   2 x  4   dx x 0 x x 2 3 dy y  10  g). f  x   lim   x3 dx x 0 x   2 x  4  2 x  4   Solution: We have dy y 10 First principle for derivative  lim  dx x 0 x  2 x  4 2 dy y f  x  x   f  x   lim  lim i). f  x   3x 2  4 x  9 dx x 0 x x 0 x 1  3  Solution: We have f  x   3x  4 x  9 2 dy 3 Substituting  lim    dx x0 x   x  x   3 x  3  First principle for derivative dy y f  x  x   f  x  dy y 1  3 3   lim  lim  lim  lim   x  x dx x 0 x x 0 x  x  x  3 x  3  dx x 0 x 0 Substituting f  x   3x  4 x  9 2 dy y 1  3  x  3  3  x  x  3   lim  lim   3  x  x 2  4  x  x   9   3 x 2  4 x  9     dx x0 x x0 x   x  x  3 x  3  dy  lim  dx x 0 x dy  lim y  lim 1  3x  9  3 x  3x  9    dy  lim  2  3 x 2   x   2 xx  4 x  4x  9  3x 2  4 x  9 dx x 0 x x 0 x   x  x  3 x  3  dx x 0 x 3x 2  3  x   6 xx  4 x  4x  9  3x 2  4 x  9 2 dy y 1  3x  dy  lim  lim  lim   x dx x 0 x x 0 x   x  x  3 x  3  dx x 0 3  x   6 xx  4x 2 dy dy y  3   lim  lim  lim   dx x 0 x x   x  x  3 x  3  dx   x 3x  6 x  4 x 0 x 0 dy y  3   lim dy  lim   dx x 0 x dx x 0 x   x  0  3 x  3  dy  lim 3x  6 x  4 dy y  3  dx x 0  lim   dx x 0 x   x  3 x  3  dy  3  0   6 x  4  dy y 3 dx  lim  dy dx x 0 x  x  32  6x  4 dx 5 h). f  x  Q2. Estimate the slope of the tangent line on a curve at 2x  4 a point P(x,y) for each of the following graph Solution: We have f  x   5 a). Solution: From graph points are  5,3  and  7, 7  2x  4 73 4 First principle for derivative slope  m   2 75 2 y f  x  x   f  x  dy  lim  lim b). Solution: From graph points are  2, 2  and  0, 4  dx x0 x x0 x 42 2 5 slope  m    1 Substituting f  x   0  2 2 2x  4 c).Solution: From graph points are  2, 2  and  4, 3  dy y 1  5 5  3 2 1  lim  lim    slope  m   dx x0 x x0 x  2  x  x   4 2 x  4  4   2  6 dy y 1  5 5  Q3a). Draw a tangent line to the graph of function  lim  lim  f  x   x 2  7 x  6 at point 1, 0  of function and dx x 0 x x 0 x  2 x  2x  4 2 x  4   Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 20  Exercise 2.2 Chapter 2 estimate it slope at the same point graphically. y  4  3x 15 Sol: since tangent line passing through one point of curve y  3x  19 …………………………………(2) therefore draw a tangent line on the curve at point 1, 0  Which passing through estimated point  3, 10  i.e. 10  3  3  19 10  9  19 10  10 Name of Rule Leibniz Notation Constant d dx c  0 dx  cf   c dx Constant d d f multiple Which line passing through the point  3, 10  Sum d dx  f  g   dxd f  dxd g dx  f  g   dx f  dx g b). Draw a tangent line to the graph of the function Difference d d d f  x   x  7 x  6 at point  5, 4  of the 2 dx  af  bg   a dx f  b dx g Linearity d d d function and estimate it slope at same point graphically. dx  fg   f dx g  g dx f Sol: since tangent line passing through one point of curve Product d d d therefore draw a tangent line on the curve at point 1, 0  f  f Quotient   g d d d f dx dx g dx g g2 Constant Rule Let f  x   c then f  x  x   c First principle for derivative d f  x  x   f  x  f  x  f  x   lim dx x 0 x Putting the value d cc f  x  f  x   lim 0 Which line passing through the point  3, 10  dx x 0 x c). Is your estimate about tangent line at the point  f  x  0 1, 0   5, 4  equal to actual derivatives of the and Constant multiple First principle for derivative function at point 1, 0  and  5, 4  cf  x  x   cf  x  Solution: we have the function f  x   x  7 x  6 2 d dx cf  x   lim x 0 x Differentiating with respect to x Putting the value f  x  x 7 2 x f  x  x   f  x  d cf  x   c lim d d d d dx dx dx dx 6 f   x   2x  7 dx x 0 x Slope of function at point 1, 0  i.e. x  1 & y  0 d dx cf  x   c d dx f  x f  1  2 1  7 Sum Rule Let h  x   f  x   g  x  then m1  5 h  x  x   f  x  x   g  x  x  So the equation of tangent y  y1  m1  x  x1  First principle for derivative y  0  5  x  1 d h  x  x   h  x  h  x   h  x   lim dx x 0 x y  5x  5 …………………………………….(1) Putting the value of h  x   f  x   g  x  Which passing through estimated point  3, 10   f  x  x   g  x  x    f  x   g  x  i.e. 10  5  3  5 d dx  f  x   g  x   lim x 0 x 10  15  5 f  x  x   f  x   g  x  x   g  x  d dx  f  x   g  x   lim x 0 x 10  10 f  x  x   f  x  g  x  x   g  x  Slope of f  x  at point  5, 4  i.e. x  5 & y  4 d dx  f  x   g  x   lim x 0 x  x f   5  2  5  7 d  f  x   g  x   dx f  x   dx g  x  d d dx m2  3 Difference Rule Let h  x   f  x   g  x  then So the equation of tangent y  y1  m2  x  x1  h  x  x   f  x  x   g  x  x  y   4   3  x  5  First principle for derivative Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 21  Exercise 2.3 Chapter 2 d h  x  x   h  x  1  g  x   f  x  x   f  x   f  x  g  x  x   g  x   h  x   h  x   lim h  x   lim   dx x 0 x x 0 x  g  x  x  g  x   Putting the value of h  x   f  x   g  x   g  x  f  x xx f  x   f  x  g  x xx g  x   h  x   lim    f  x  x   g  x  x    f  x   g  x  x 0  g  x  x  g  x   d  f  x   g  x   lim  x dx  g  x  lim f  x x   f  x   f  x  lim g  x x   g  x   x 0 f  x  x   f  x   g  x  x   g  x  h  x    x 0 x x 0 x  d  f  x   g  x   lim x  lim g  x  x  g  x   dx x 0   x 0  f  x  x     g  x  x   g  x   f x d  f  x    g  x  dx f  x   f  x  dx g  x  d  f  x   g  x   lim d d x x    dx  g  x    g  x g  x dx x 0  d  f  x   g  x   dx f  x   dx g  x  d d d f  x   g  x  dx f  x   f  x  dx g  x  d d dx   dx  g  x   Linearity Rule Let h  x   af  x   bg  x  then  g  x   2 h  x  x   af  x  x   bg  x  x  Exercise 2.3 Q1. Use the product rule to find out the derivative First principle for derivative of following functions; d h  x  x   h  x  h  x   dx h  x   lim x 0 x a). y   x 2  2   3 x  1 Putting the value of h  x   af  x   bg  x  af  x  x   bg  x  x   af  x   bg  x  Sol: Differentiating dy d  dx dx  x 2  2   3x  1  d dx af  x   bg  x   lim x 0 x Using the product rule af  x  x   af  x   bg  x  x   bg  x    x 2  2   3 x  1   3 x  1  x 2  2  dy d d d af  x   bg  x   lim  x dx x 0 dx dx dx f  x  x   f  x  g  x  x   g  x  d dx af  x   bg  x   a lim x 0 x b x dy    x2  2  3 d d  x  1   3 x  1   d 2 d  x  2 dx  dx dx   dx dx  d dx af  x   bg  x   a d dx d f  x  b g  x dx y   x 2  2   3 1  0    3x  1 2 x  0  Product Rule Let h  x   f  x .g  x  then y  3  x 2  2   2 x  3x  1 h  x  x   f  x  x  g  x  x  y  3x 2  6  6 x 2  2 x First principle for derivative y  9 x 2  2 x  6 d h  x  x   h  x  h  x   h  x   lim dx x 0 x b). y   6 x3  2   5 x  3 Putting the value of h  x   f  x .g  x  d h  x   lim  f  x  x  g  x  x   f  x  g  x  1 Sol: Differentiating dy d  dx dx   6 x 3  2   5 x  3  dx  x 0 x Using the product rule 1  f  x  x  g  x  x   g  x  x  f  x   dy   6 x 3  2   5 x  3   5 x  3  6 x 3  2  d d h  x   lim   x 0 x    g  x  x  f  x   f  x  g  x    dx dx dx  d d   d 3 d    6 x3  2   5 x  3    5 x  3  6 dy 1  g  x  x   f  x  x   f  x    x  2 h  x   lim   dx  dx dx   dx dx  x 0 x             f x g x x g x f  x x  f  x  g  x x  g  x   y   6 x3  2   5 1  0    5 x  3 6  3 x 2   0  h  x   lim g  x  x  lim x  f  x  lim x y  5  6 x3  2   18 x 2  5 x  3 x 0 x 0 x 0 y  30 x3  10  90 x3  54 x 2 d  f  x  g  x   g  x  f  x   f  x  g  x  d d dx dx dx y  120x3  54x2  10 Quotient RuleLet h  x   g  x  , g  x   0 then f  x c). y   7 x 4  2 x  x 2  4  f  x x  h  x  x   g  x x  Sol: Differentiating dy  d dx dx  7 x 4  2 x  x 2  4   First principle for derivative Using the product rule d h  x  x   h  x  h  x   h  x   lim dy   7 x4  2x   x2  4   x2  4  7 x4  2x  d d dx x 0 x dx dx dx f  x  d 2 d   d 4 d  Putting the value of h  x   g  x    7 x4  2x   4    x2  4  7 dy x  x 2 x dx  dx dx   dx dx  d  f  x .g  x   lim 1  f  x  x  f  x       y   7 x 4  2 x   2 x  0    x 2  4  7  4 x 3   2 1  dx  x  g  x  x  g  x   y  2 x  7 x 4  2 x    x 2  4  28 x 3  2  x 0 d 1  g  x  f  x  x   f  x  g  x  x   h  x   lim   y  14x5  4x2  28x5  2x2 112x3  8 dx x 0 x  g  x  x  g  x   1  g  x  f  x x  g  x  f  x  f  x  g  x x  g  x  f  x   y  14x5  28x5 112x3  4x2  2x2  8 h  x   lim x 0 x  g  x x  g  x   y  42x5 112x3  6x2  8 Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 22  Exercise 2.3 Chapter 2 d). y   2 x 2  4 x  3 5 x3  x  2  dy  1   4 1 1    3 x 2  6   x 2  0  dx   2  Sol: Differentiating dy  d dx dx  2x 2  4 x  3 5 x3  x  2   1   3 12 1  Using the product rule   4x 2  2   x  0   2    2 x 2  4 x  3  5 x3  x  2  dy d dy 1  1  3 1  1  dx dx  2 x 2  3x 2  6   x 2  4 x 2  2    5 x3  x  2  dx   2   d dx  2 x 2  4 x  3 1 1 1 1 1 1 dy  3  3  d 3 d d   6 x 2 2  12 x 2   4 x 2 2   2 x 2   2 x 2  4 x  3  5 dy x  x 2 dx 2 2 dx  dx dx dx  1 1 dy  d 2   5 x3  x  2   2 x 4 d x d  3  6 x0  12 x 2  6 x 0  3x 2  dx dx dx  dx 1 1  y   2 x 2  4 x  3 5  3 x 2   1  0  dy dx  12 x 2  3x 2  6  6   5 x3  x  2   2  2 x   4 1  0  dy 1  15 x 2  12 y   2 x 2  4 x  315 x 2  1   5 x3  x  2   4 x  4  dx Q2. Use the quotient rule to find out the derivative y  30 x 4  2 x 2  60 x3  4 x  45 x 2  3 of the following functions; 20 x 4  20 x3  4 x 2  4 x  8 x  8 3x  5 a). y y  30 x 4  20 x 4  60 x3  20 x3  2 x 2  45 x 2  4 x 2 x4 4 x  4 x  8 x  8  3 dy d  3 x  5  Solution: Differentiating    dx dx  x  4  y  50 x 4  80 x3  39 x 2  16 x  5 Using Quotient rule e). y   2 x  3  x 1  dy  x  4  dxd  3x  5   3x  5 dxd  x  4    x  4 2 Sol: Differentiating dy  d    1   2 x  3  x 2  1  dx dy  x  4   3 dxd x  dxd 5    3 x  5   dxd x  dxd 4  dx dx      Using the product rule   x  4 2 dx dy d  12   12  d   2 x  3     x  1  2 x  3  dy  x  4   3 1  0    3 x  5 1  0  x 1 dx dx     dx   x  4 2 dy  d 12 d   12  d d  dx   2 x  3  x  1   x  1  2 x 3 dx  dx dx     dx dx  dy 3  x  4   1 3x  5    1 1 1   1   x  4 2 dx   2 x  3  x 2  0    x 2  1  2 1  0  dy dx 2    dy 3x  12  3x  5     x  4 1 1 2 dy 1 2 dx  x  2 x  3  2  x 2  1 dx 2   dy 7 1 1 3 1  dx  x  4 2 1 dy  x 2  2x  x 2  2x 2  2 dx 2 2 1 1 2 y 1 dy 1 3 b).  x 2  x 2  2x 2  2 3x  5 dx 2 1 1 3 1 dy d  2  dy  x 2  2x 2  x 2  2 Solution: Differentiating    dx 2 dx dx  3 x  5  1 1 Using Quotient rule dy 3  3x 2  x 2  2 dx 2 dy  3x  5  dtd 2  2 dtd  3x  5  f).  y  3 x  6 4 x  2     3x  5 2 dx Sol: Differentiating dy  d dx dx  3 x 6  4 x 2  dy  3x  5 .0  2  3 dtd x  dtd 5  Using the product rule   3x  5 2 dy  1  d  1  dx   3 x 2  6   4x  2  2     dy 2  3 1  0  dx dx  1  d  1     4x 2  2   3 x  6   3x  5 2 2   dx   dx     6 1 1 dy   3 x 2  6   4 d x2  d dy 2     dx dx  3 x  5  2 dx dx   1  d 1 d    4 x 2  2   3 x2  6   dx dx  Khalid Mehmood Lect: GDC Shah Essa Bilot Available at http://www.MathCity.org Page 23  Exercise 2.3 Chapter 2 c). f t   t2  t dy  1   x d  5 x  6    5 x  6  dxd x t 1  x   2 dx dx d d t 2  t  f t     dy 1    x  5 Solution: Differentiating 1 dt dt  t  1    d dx x d dx 6  5x  6 x  d dx 2 Using Quotient rule dx x    t  1 dtd  t 2  t    t 2  t  dtd  t  1 dy 1  1    1 x  5 1  0    5 x  6  x 2  d 1 f t      t  1 dx x   2 dt 2  t  1  dtd t 2  dtd t    t 2  t   dtd t  dtd 1 dy 1  1 1  f  t    5 x   5 x  6  x 2   t  1 2 dx x  2   t  1 2t  1   t 2  t  1  0  dy 1   5 x  5x  6  f  t   dx x    t  1 2 2 x  dy 1 10 x   5 x  6    t  1 2t  1  1 t 2  t     f  t   dx x  2 x   t  1 2 dy 1  5 x  6  2t 2  t  2t  1  t 2  t  f  t   dx x  2 x   t  1 2 dy 5 x  6 2t 2  t 2  t  t  2t  1  f  t   dx 2 x x  t  1 2 x2  7 x  2 t  2t  1 2 f). y f  t   x2  t  1 2 dy d  x 2  7 x  2  Solution: Differentiating    x  6x 2 dx dx  x  2  d). y 4 x3  1 Using Quotient rule dy d   x 2  6 x  dy  x  2  dx  x  7 x  2    x  7 x  2  dx  x  2  d 2 2 d Solution: Differentiating     dx dx  4 x3  1  dx  x  2 2 dy  x  2   dx x  7 dx x  dx 2    x  7 x  2   dx x  dx 2  Using Quotient rule d 2 d d 2 d d dy      4 x3  1 dxd  x 2  6 x   x 2  6 x   4x d 3  1    x  2 dx 2 dx   2 dx 4 x3  1  x  2   2 x  7 1  0    x 2  7 x  2  1  0        dy dy 4 x  1  dxd x  6 dxd x   x  6 x 4 dxd x  dxd 1 3 2 2  3  x  2 2  dx   2 dx 4 x3  1 dy  x  2  2 x  7   1 x  7 x  2  2 dy  4 x  1  2 x  6 1     x  6 x  4  3x   0  3 2 2 dx     x  2 2  4 x3  1 2 dx dy 2 x 2  7 x  4 x  14  x 2  7 x  2  dy  4 x  1  2 x  6   12 x   x  6 x   x  2