RI Differentiation C6A Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 1 of 8 C6A: Differentiation Techniques (Additional Practice) 1 Differentiate each of the following with respect to x: (a) 22(1 ) e xx (b) cos 4x x   (c) 1 1s i n x (d) 12ln tan ( ) xx     (e) 2 2 1ln 1 x x   (f) lnexx (g) 1tan 3e x (h) 23sec 2xx (i) cot cos ec 2xx (j) 1sin (tan )x (k) 21cos (e )xx  (l) xsin x Solution: (a)        22 22 2 2 2 d 1e 2 1 e 1e 2d 21 e 1 1 21 e xx x x x xx xx xx xx           (b) 1 2 2 3 1sin coscos( ) 44 2d 4 d 1 2s i n c o s 442( ) xx x xx x x x xx x x                     

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 2 of 8 (c) 2 2 d1 (1 sin ) cosd1 s i n cos = (1 sin ) x xxx x x       (d)  1 2 11 2 1 1 d2 1 2 2ln tan tan lnd 1 11 2tan ln 21 x xxxx x x x xx xxx                      (e) 2 22 2 22 4 d1 1 dln [ln(1 ) ln(1 )]d1 2 d 12 ( 2 ) 21 1 2 1 x x xxxx xx xx x x         (f)  ln ln ln d1(e ) e 1 lnd e1 l n xx xx xx xxx x x     (g)  11 1 tan 3 tan 3 2 tan 3 2 d1[e ] e . 3d 13 3 e 13 xx x x x x        (h)  23 3 2 2 3 d ( sec 2 ) 2 sec 2 (3sec 2 )(sec 2 tan 2 )(2)d 2 sec 2 1 3 tan 2 xx x x x x x xx xx x x   (i)   2 2 d (cot cosec2 ) cosec cosec2 cot cosec2 cot 2 (2)d cosec2 cosec 2cot cot 2 xx xx x x xx xx x x      (j) 2 1 2 ds e c[sin (tan )]d 1t a n xxx x    (k) 21 1 2 2 1 2 d1[ cos (e )] 2 cos (e ) .ed 1( e ) e 2cos (e ) 1e x xx x x x x xx xx xx           (l) Let sin xyx .

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 3 of 8 Taking ln on both sides, ln (sin ) lnyx x . Differentiate with respect to x, sin 1d s i n(cos )(ln )d ds i n (cos )(ln )d x yx xxyx x yx x xxxx    Alternatively, sinln (sin )lnee xx xxy (sin )ln sind1 s i ne (cos ) ln (sin )( ) (cos ) lnd xx xy xxx x x xxx xx           2 Find 21d 99 s i nd3 xxxx     , leaving your answer in the form 2ab x , where the values of a and b are to be found. Solution: 1 21 2 2 2 2 2 22 2 d1 9 9 9sin ( ) 9 ( )(9 ) ( 2 )d3 2 9 992 9 9 () xxx x x x xx x xxx x           a = 2, b = 9. 3 For each of the following, find d d y x in terms of x and y. (a) 33 230xyx y (b) tan 0xx y (c) cos( ) ln( ) 0xy x y  Solution: (a) 33 230xyx y Differentiating w.r.t. x, 22 22 2 2 dd36 3 3 0 dd d(6 3 ) 3 3 d d d2 yyxy x y xx yyx xy x yx y x xy        

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 4 of 8 (b)  tan 0xx y Differentiating w.r.t. x,      2 22 2 2 d1s e c 0 d dsec 1 sec d 1s e cd ds e c yxy x y x yx xy y xyx yx yy x xx y       (c) cos( ) ln( ) 0xy x y  Differentiating w.r.t. x, d1d dsin( ) 0 d dd() s i n ( ) () s i n ( ) 1 0dd d1 ( ) s i n ( ) d 1 ( )sin( ) y y xxy x y xx y yyxx y x y yx y x y xx yy x y x y x xx y x y            4 JPJC Promo 9758/2021/Q3 It is given that 21 3tan 4xy y  . Find d d y x in terms of x and y. [4] Hence, find the exact value of d d y x when y = 1, given that x > 0. [3] Solution: 21 3tan 4xy y  Differentiate with respect to x,  2 2 d1 d 20d1 d yyxy xxy x     2 2 d1 2d1 y x xyxy    2 2 d2 d 1 1 yx y x xy       2 22 2 21 1 xyy x xy    . when y = 1, 21 3(1) tan 1 4x   2 3 44x    2x  x  . Since x > 0, x  .   2 2 21 1d d1 ( 1 ) y x      = 4 12   (exact).

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 5 of 8 5 SRJC/2007/1/Q5 (a) Find d d y x if (i) 3 21y x  (ii) 2 1ln 1c o s xxy x     (b) Given that 3e(5 )yyx y , find d d y x in terms of x and y. Solution: (a) (i) 1 23 3(2 1 )21yx x     3 2d3 (2 1) (2)d2 y xx    = 3 2 3 (2 1)x   (ii) 2 21ln ln( 1) ln(1 cos )1c o s xxyx x x x        2 d2 1s i n d1 1 c o s yx x x xx x    (b) 3e(5 )yyx y  2 22 2 2 dd dee 3 ( 5 ) 1 5dd d d1 e 1 5 (5 ) 3 (5 ) d d3 ( 5 ) d( 1 ) e 1 5 ( 5 ) yy y y yy y yx yxx x yyx y x y x yx y x yx y          6 ACJC Promo 9758/2021/Q3 (modified) A curve C has equation 22 22 41 100 2 xy xx y   , x  , 8.x Show that 2d2 d2 1 6 yx y x xy y   . [2] Solution: 22 222 8 100xy x x y  . Differentiate with respect to x, 2dd41 6 2 2 dd yyxy x x y yxx   2d2 d2 1 6 yx y x xy y   (shown).

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 6 of 8 7 If sin 2sinyx , show that . Solution: If sin 2sinyx , differentiating both sides with respect to x, dcos 2cosd d 2cos secd yyx x y x yx   Squaring both sides, 2 22 2 2 2 2 2 2 2 2 22 2 2 2 d4 s i n( ) 4cos sec 4(1 sin )sec 4secdc o s sin4sec , since 2sin sincos 4sec tan 4sec (sec 1) 13 s e c ( S h o w n ) yx xy x y yx y yyx y y yy y y y           8 CJC Promo 9740/2013/Q4 (a) Given that 1tan ,yx  find d d y x . [2] (b) Given that ඥ𝑦ೣ ൌ √𝑥 ೤ , where 𝑥൐ 0, 𝑦൐ 0, find d d y x . [4] Solution: (a)  1 1 2 2 d1 1 1tand2 1( ) 2 ( 1 ) xxx x xx     (b) yx yx Taking logarithm on both sides, 11ln ln ln ln yxxy yyxx   Differentiating both sides,  1d d 1 ln lndd d1l n 1l nd d1 l n d1 l n yyyy x xyx x x yyx x yx xy         d d 1+3 s e c 2 2y x y    

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________ Additional Practice C6A: Differentiation Techniques Page 7 of 8 9 Given the curve with equation 22 431 0xx y y  , find the gradient of the curve at the point where the y coordinate is 1. Solution: Differentiate 22 431 0xx y y  with respect to x , dd2( 44 ) 6 0 dd yyxy x y xx   --- (1) When 1y , the equation of the curve gives 2 44 0xx  , so 2x . Put the values into (1) to get dd4 4 4(2) 6 0 dd yy xx    . So dd20 0dd yy xx   . Hence the gradient at (2, 1) is 0. 10 (a) Find an expression for d d y x in terms of t for 2 2 22 88, 2 txt y tt  , where 0t  . (b) DHS Mid Year 9758/2021/Q8 (modified) A curve C has parametric equations 2 2 πs i n2 , c o s 1 , 0 .xt yt t    Express d d y x in the form tan ,ab t where a and b are constants to be determined. [2] (c) MJC Promo 9