05_2016_-_2017_H2_Maths_Differentiation_Techniques_Assignment_Solutions_(LMS)

National Junior College Mathematics Department 2016 Differentiation Techniques Page 1 of 1 National Junior College 2016 – 2017 H2 Mathematics Differentiation Techniques Assignment Solutions 1(a)   2d ln sind xxx      2cos 2 ln sin 2 [M1]sin 2 2 cot 2 ln sin 2 . [A1] xxx x x x x   (b)   2 tand ed kxxx       2 2 tan tan tan 2 sec e 2 e [M1] e sec 2 . [A1] kx kx kx x k kx x x kx kx      (c) 1 2 d sin d1 x xx            1 2 222 12 2222 12 22 2 sin 1 1 [M1]111 2 sin 1 11 2 sin 1 . [A1] 1 xx xxx x x x xx x x x x           2(a) cosyxx . Taking “ln” on both sides, we get:  ln ln cosy x x  . [M1] Differentiate implicitly wrt x , we get:     2 d sinln [M1]d cos d ln tand ln cosd tan . [A1]d ln ln y y x xx x x yy xxxx xyx xx xx        2(b) 1 ey x yx  . Taking “ln” on both sides, we get ( 1) ln . [M1]y x x y   Differentiate implicitly wrt x, we get   d 1 dln 1 [M1]dd d1ln 1 1 d d1 . [A1]d (ln 1) y y yx x x x yyx xx y x y x x x         3. d 2 2 12,d2 xt t t t      2d 2 22 1 . [M1]d y ttt t t        2 2 d 2 1d d [A1]21dd d 21 .21 y ty t t txx tt t t        Given that d 2d y x  ,     2 2 2 21 2 [M1]21 1 2 1 20 20 t t tt tt tt        0 or 2tt   Since 0t , 2. [A1]t 