TMJC H2 Chapter 7 Differentiation Extra Practice Solutions

Chapter 7 Differentiation TMJC 2024 Page 1 of 6 H2 Mathematics (9758) Chapter 7 Differentiation Extra Practice Solutions Q1 Solutions (a) Differentiate wrt x, 4 2 3 43 3 4 dd( 1) (1) 2 (4 ) 0dd d ( 1) 2 (1 4 )d d (1 4 ) d ( 1) 2 yyx y x y y xxx y x yx y x yx y y x y x x x y + + + + =  + + =− + −+= ++ (b) , x > 0 xyxx xyx x ln2ln ln2elnln =+ =+ Differentiate wrt x, xx yx x y x yx x yx x yxxyx ln2 21 d d 21 d dln2 d d)(ln21211 −+= −+= +    =+ (c) Differentiate wrt x, 22 22 22 22 22 2222 22 e2e2 e2 d d e2)e2e2(d d d d d d22ed d2e −+ + +−+ −+ +− = =+−     −=     +−     yy x x y xyyx y x yyx yyxx y yxy yx yxyxy yxy (d) Differentiate wrt x, 42( 1) 1x y x y+ + = 2ln e lnxx y x= 222 1eey x y y +−= ( ) ( ) 2 2 cos 3y x y xyx + − − =

Chapter 7 Differentiation TMJC 2024 Page 2 of 6 )sin()(22 )sin()(2 d d )sin()(2)sin()(22 d d 0)sin(d d)sin(d d)(2)(2d d2 0d d))sin((d d1)(2d d2 2 2 2 2 2 xyxyxx y xyyyxx y x y xyyyxx yxyxyxx y x y xyyx yxyxx yyxyxx y x y x y yx yxxyx yyx x yx yyx +−− −−−     = −−−    =     +−− =++−−−+    − =     +−−     −−+ −     Q2 Solutions 22 54 0x y xy+ + = Differentiating w.r.t x: 22 2 2 dd 2 2 0dd d 2 (2 ) d 2 ( 2 ) yyx xy y xyxx y xy y y x y x x xy x x y + + + = ++=− =−++ Gradient 1=− : 2 2 2 12 xy y x xy +− =−+ 22 22 22 or xy y x xy yx y x y x + = + = = =− 33 When : 54 0 (contradiction reject ) yx x x y x =− − + + =  =− 33 3 When : 54 0 27 3 yx xx x x = + + = =− =− Hence there is only one such point ( 3, 3) .−−

Chapter 7 Differentiation TMJC 2024 Page 3 of 6 Q3 Solutions (a) 11 22 d 3 9sin(cos (3 )) cos(cos (3 ))d 1 9 1 9 xxxx xx −− −− =  = −− (b) ( ) 1 2 2 d sin (cos )d 1 ( sin ) 1 (cos ) sin sin sin sin sin if sin 0sin sin if sin 0sin 1 if sin 0 1 if sin 0 xx x x x x x x x xx x xx x x − =− − =− =− − −=−  =− Q4 Solutions At point where dd 1 cos , sindd xy = − = d d d sin d d d 1 cos yy xx  =  = − ,= d1 d2 y x = sin 1 1 cos 2 2sin 1 cos 2sin cos 1 (shown)     =−  = −  + =

Chapter 7 Differentiation TMJC 2024 Page 4 of 6 Q5 Solutions (i) 22 90x xy y− + − = Differentiate wrt x : dd2 2 0 dd yyx y x y xx− − + = ( ) d22 d yy x y x x− = − ( ) d22 d yy x y x x− = − (ii) At stationary points, d 0d y x = . We get 2 0 2y x y x− =  = Substitute y = 2x to eqn of C: 2 2 22 4 9 0x x x− + − = 23 9 0x −=  3x= and hence 23y= The exact coordinates of the stationary points are ( )3, 2 3 and ( )3, 2 3−− (iii) Differentiate ( ) d22 d yy x y x x− = − wrt x : ( ) 2 2 d d d d2 1 2 2d d d d y y y y yxx x x x  − + − = − At stat points, , d 0d y x = , ( ) 2 2 d2 d2 y x y x −= − At ( )3, 2 3 , we have 2y-x > 0. ( ) 2 2 d2 0d2 y x y x −= − . Max point At ( )3, 2 3−− , we have 2y-x < 0, ( ) 2 2 d2 0d2 y x y x −= − Min point

Chapter 7 Differen