1. 2022 Techniques of Differentiation Essential Practice Solution
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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 − − = + =− + =− +
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