1. 2022 Techniques of Differentiation Essential Practice Solution

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 1 H2 MATHEMATICS TUTORIAL SOLUTION TOPIC TECHNIQUES OF DIFFERENTIATION 2022/JC1 ESSENTIAL PRACTICE 1 [2014/PJC/Promo/1] Differentiate each of the following with respect to x. (i) sece x [1] (ii) ( ) 12tan x− [2] (iii) ( )4cos 2 x [2] Solution: (i) ( ) ( )sec secd e sec tan ed xx xxx = (ii) ( ) 12 4 d2 tand 1 xxx x − = + (iii) ( ) ( ) ( )( ) ( ) ( ) 43 3 d cos 2 4cos 2 sin 2 2d 8cos 2 sin 2 x x xx xx  =− =−

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 2 2 [2013/NYJC/Promo/2] Differentiate the following expressions with respect to x, simplifying your answers as far as possible: (a) 1 2tan x −   , [3] (b) 1ln 1 x x + − . [3] Solution: (a) 1 2 2 2 d 2 2( )tand 21 2 4 x xx x x − − − =  + =− + (b) Method :   2 d 1 1 dln ln(1 ) ln(1 )d 1 2 d 1 1 1 2 1 1 11 or (1 )(1 )1 x xxx x x xx xxx  + = + − − − −=−  +− = +−− Method : [Not Recommended] ( ) ( ) ( ) 2 2 d 1 1 1 1 1 1ln .d 1 2 1 1 1 1 1 1 2 21 1 1 1 (1 ) x x x x x x x x x x x x x xx     + − − + +  = −+  + −  − − =  +  − = −+

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 3 3 [2013/ACJC/Promo/3] Differentiate the following with respect to x. (i) 1cos 2 x−   , [2] (ii) ( ) 3 2 1ln 1 x x + − . [2] Solution: (i) ( ) 1 211 2 21 d 1 1 1cos cosd 2 2 2 2 1 2 1 2 4 cos 2 xx x x xx − −− −   −   =          − =− −  (ii) Method : ( ) ( ) ( ) ( ) ( ) 3 2 2 1d d 1ln ln ( 1 1 1)dd 11 d1 ln 1 ln 1d2 11 1 2 1 x x x x xxx xx xxx xx  +  + =   + = +  −−  = + − − =− +− Method : [Not Recommended] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 23 23 22 33 2 22 2 3 1 1 1 21d 1 1 1lnd2 1 11 1 11 3 21 x x x xx x x xx x xx x x  + − − ++ =   − ++ − −− −= −

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 4 4 [2016/PJC/Promo/1] Differentiate with respect to x, giving your answers as a single fraction, (a) 22ln ax− , where a is a constant, [2] (b) 1 1tan 2x −   . [2] Solution: (a) Method : ( ) ( ) 2 2 2 2 22 22 22 d d 1ln lnd d 2 12 2 a x a xxx x ax x ax x xa − = −  −=  − =− − = − Method : [Not Recommended] ( ) ( ) 1 2 2 2 2 2 22 22 22 d 1 1ln . .( 2 )d2 = a x a x xx ax x ax x xa − − = − − − − − = − (b) 1 22 2 2 2 d 1 1 ( 2)tand2 (2 )11 2 12 1 41 4 2 41 xx x x x x x − − = +   =−   + =− +

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 5 5 [2015/CJC/Promo/1] Differentiate the following expressions with respect to x , simplifying your answers whenever possible. (a) ( ) 13tan e x− , [2] (b) 25 x , [2] (c) 3 1 ax x + , where a is a constant. [3] Solution: (a) ( ) ( ) ( ) 1 3 3 23 3 6 d 1 dtan e edd 1e 3e 1e xx x x x xx − = + = + (b) Method : Let 25 xy= ( ) ( )( ) 2 2 d 5 ln 5 2d 2ln 5 5 x x y x =   = Method : Logarithmic Differentiation Let 25 xy= Take ln on both sides: ( ) ( ) ( )( ) 2 2 ln ln 5 ln 2 ln 5 1d 2ln 5d d 2ln 5d 2ln 5 5 x x y yx y yx y yx = = = = =

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 6 (c) Method : Product Rule ( ) ( ) ( ) ( ) ( ) ( ) 1 3 2 3 13 3 3 2 22 13 3 3 3 22 dd 1dd 1 11 1 3 2 31 1 2 ax ax xxx x a x ax x x a x ax x − −− −−  =+ +  = + + − +   = + − + Method : Quotient Rule ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 3 3 2 2 23 3 11 3 3 3 22 3 13 3 3 3 22 11 1 3d 2 d 1 1 311 2 1 31 1 2 a x ax x xax x x x a x ax x x a x ax x − − −− + −  + = + + + − + = + = + − +

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 7 6 [2010/ACJC/Promo/5] Differentiate the following with respect to x , leaving your answers in terms of x . (a) ee cot 2 x x  , [3] (b) ( )( )1/2 xx . [4] Solution: (a) ( )( )e e 2 e e2 d1 e cot e cosec cot e ed 2 2 2 2 1e cosec ecot2 2 2 x x x x x x x x xx           = − +                         = − +         (b) Let ( ) 1 2 xyx= ( ) ( ) ( ) ( ) 2 1 2 1ln ln 2 1 d 1 2 1 ln 2d2 2d 1 ln 2d x yx x y xy x x x x xy xx x = =−  =  − 

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 8 7 [2012/Promo/ACJC/2] Differentiate the following with respect to x. (i) ( )1cos sin x− , where π 3π 22 x , [2] (ii) e1ln 1e x x− + − . [3] Solution: (i) ( ) ( ) ( ) ( ) 1 2 2 2 d1 cos sin cosd 1 sin cos cos cos π 3π cos 0 when so cos coscos 2 2 1 xxx x x x x x x x xx − =−  − −= − =    =−−  = (ii) d e 1 d 1 e 1ln lnd d 2 1 e 1 e xx xxxx −−   ++ =   −−   ( ) ( ) 1d ln e 1 ln 1 e2d 1 e e 2 e 1 1 e 1 e 1 2 e 1 e 1 xx xx xx x xx x − − − = + − − =−  +− =−  +− _________________________________________________________________________________

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 9 8 [2016/CJC/Promo/1] Given that n3 ta πe cos 2 3 yyx + = +  , find d d y x in terms of x and y . [4] Solution: ( ) t 3 22 22 22 tan tan tan an e cos 2 3 dd3 sec e 2sin 2d d 3 d3 sec e 2sin 2 d3 2sin 2d 3 d 3s π π ec π π e y y y y yx yyy y xxx yy y x x xy x yy + = +  +   =− +  +  =− +   +=− + 9 [2014/CJC/Promo/2] The variables x and y are related by 3 secy xy x=− . Find d d y x in terms of x and y . [4] Solution: ( ) 3 sec dd3 ln 3 sec tandd d3 ln 3 sec tand d sec tan d 3 ln 3 y y y y xy x yy y x x xxx yx y x xx y y x x xx =− = + − − = − −= −

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: TECHNIQUES OF DIFFERENTIATION Page | 10 10 [2012/Promo/CJC/11(b)] Given that ( )21 x yx=+ for 0x , find d d y x . [3] Solution: ( ) ( ) ( ) ( ) ( ) ( ) 21 ln ln 2 1 1 d 2 ln 2 1d 2 1 d   2 ln 2 1d 2 1 22 1 ln 2 121 x x yx y x x y xxy x x yx yxxx xxx x =+ =+  = + +  + = + + + = + + +  + 11 [2013/YJC/Promo/4a] It is given that 2exyyx= . Find d d y x in terms of x and y , simplifying your answer. [4] Solution: ( ) ( ) ( ) 2 2 e ln ln e ln 2ln 1 d d 2ln dd d21 lnd 2 lnd d 1 2 ln 2 ln xy y yx x y x x y y x yyyx y x x x xy yy x x yy x xx y xy x xy y y x y x x y = = =+ + = +  − = − − = − − = − −= −