TMJC H2 Chapter 7 Differentiation Discussion Questions Solutions 2024

Chapter 7 Differentiation TMJC 2024 Page 1 of 19 H2 Mathematics (9758) Chapter 7 Differentiation Discussion Solutions Level 1 1 Differentiate the following with respect to the variable given: (a)  cotx x (b)  3ln( )z z (c)   3 log 1 3a x , where a is a positive constant and 1a (d)   2ln sin  (e) sin3x x (f) 1 1 cos 2 x (g) 3sin x (h) 1cosec 3 2x     (i)   1 3tan 1 x  Solutions: (a)    d cotd x xx         2 2 1 1cot cosec 2 2 1 1cot cosec 22 x x x x x x x x      (b)    3d ln( )dz z z         2 3 2 3 13 ln( ) ln( ) 3 ln( ) ln( ) z z z z z z         ALERT:    3ln( ) 3lnz z  (c) First, we do a change of base for   3 log 1 3a x .         3 3 3 e e log 1 3 log 1 3 ln 1 3 3ln 1 3 log ln ln a x x x x a a a          3d log 1 3d a xx  Always simplify ln expression before differentiating Note that a is a constant, 0, 1a a 

Chapter 7 Differentiation TMJC 2024 Page 2 of 19      ln 1 3d3 d ln 3 3 ln 1 3 9 ln 3 1 x x a a x a x            (d)    2d ln sind  2 1 (2sin cos ) sin cos2 sin 2cot          (e) Method 1: Using Implicit Differentiation sin3x x Let sin3x xy ln sin ln 3y x x       sin Differentiate wrt , 1 d cos sin ln3d d (ln3) cos sind d 3 (ln3) cos sind x x x y x x xy x y y x x xx y x x xx       Method 2: By memorising formula   d sin3d x x x   sin3 ln 3 cos sinx x x x x  Note: The power is a variable in x, so we will need to use implicit differentiation here. ALERT: You cannot do the usual way     d sin sin 13 sin 3dx x x x x x x  Replace y by sin3x x Lecture Notes pg 4:   f ( ) f ( )d f ' lnd x xa a x ax 

Chapter 7 Differentiation TMJC 2024 Page 3 of 19 (f) 1 d 1 d cos 2x x                     11 21 2 21 2 21 2 21 2 d cos 2d 11 cos 2 1 2 2 cos 2 1 2 2 cos 2 1 2 (2 ) 2 cos 2 1 4 xx x x x x x x x x                       (g)     3 2 d sind 3sin cos xx x x (h) 1 2 d 1 cosec 3d 2 1 1 1 1cosec 3 cosec 3 cot 3 (3)2 2 2 2 3 1 1cot 3 cosec 32 2 2 xx x x x x x                                                        (i)        1 3 2 23 2 3 6 d tan 1d 1 3 1 1 3 2 2 xx x x x x x          ALERT: 1 1cos cosx x   Therefore, since 1 1 1cos 2 cos 2 1 cos 2cos 2 x x xx     inverse of cosine function reciprocal of cosine function MF26: Note: Since this question is to differentiate 1cos 2 x , remember to chain rule 2x, i.e. differentiate 2x to get 2.  f x  f ' x

Chapter 7 Differentiation TMJC 2024 Page 4 of 19 2 2004/I/14 modified Find the x-coordinates of all the stationary points on the curve   3 2 xy x a   where 1a and state, with reasons, the nature of each point. [7] 2 Solutions   3 2 , 1xy a x a                     2 2 3 4 4 2 2 3 3 2d d 3 2 3 x a x x a x x a xy x x a x a x x a x x a x a              At stationary points,     2 3 3d 0d x x ay x x a        2 2 3 0 0 or 3 0 x x a x x a      0 or 3x x a    To determine the nature of the stationary points, we will use the first derivative test. The reason why we do not recommend the use of second derivate test here is because it is very tedious to find 2 2 d .d y x x 0 0 0 d d y x  At 0x  , it is a stationary point of inflexion x  3a   −3a  3a   d d y x  At 3x a , it is a maximum turning point. Always simplify by factorising, in the case for the numerator, the common factors are    2x a x . Since 1a  , for  3x a    , we can choose a value like 2x a and substitute into       2 3 3 d 2 2 2 d 0a a a ay x a      Since 1a , for  3x a    , we can choose a value like 4x a and substitute into       2 3 3 d 4 4 4 d 0a a a ay x a     

Chapter 7 Differentiation TMJC 2024 Page 5 of 19 Level 2 3 Prove that 1 2 d 1(tan ) d 1 x x x    , where 1tan 2 2 x     . Let y = tan–1 x where 1tan2 2 x     . Differentiating with respect to x, d d(tan ) ( )d d y xx x  2 dsec 1 d yy x  2 d 1 d sec y x y 2 d 1 d 1 tan y x y  2 d 1 d 1 y x x  4 2017(9758)/I/5 When the polynomial 3 2x ax bx c   is divided by  1x ,  2x and  3x , the remainders are 8, 12 and 25 respectively. (i) Find the values of a, b and c. [4] A curve has equation  fy x , where   3 2 f x x ax bx c    , with the values of a, b and c found in part (i). (ii) Show that the gradient of the curve is always positive. Hence explain why the equation  f 0x  has only one real root and find this root. [3] (iii) Find the x-coordinates of the points where the tangent to the curve is parallel to the line 2 3y x  . [3] Solutions (i) 7 4 2 4 9 3 2 a b c a b c a b c          By GC, 3 3, and 7.2 2a b c   Learn your GC keystrokes well: Recall your A math: Reminder and Factor Theorem. There is NO NEED to do any Long Division here!

Chapter 7 Differentiation TMJC 2024 Page 6 of 19 (ii)   3 2 2 2 3 3f ( ) 7 2 2 3 1 3f ' 3 3 3 0 for all 2 2 4 x x x x x x x x x                 Method 2: Using Discriminant and coefficient of 2x :     2 33 4 3 9 2D         and coefficient of 2 3 0x   ,  f ' 0x  for all x  . Therefore the gradient of the curve is always positive. (shown) Since there are no turning points and  f x is a polynomial, the curve cuts the x-axis exactly once. Therefore  f 0x  has exactly one real root. From GC, the root is 1.33 (3 s.f.). (iii) When the tangent to the curve is parallel to the line 2 3y x  .  f ' 2x  2 2 33 3 2 2 13 3 0 2 x x x x       Using GC,  1.15 or 0.145 3 s.fx  When trying to show that gradient is positive, you need to use algebraic method. Method 1: completing the square since  f ' x is a quadratic expression. Gradient of line is 2.

Chapter 7 Differentiation TMJC 2024 Page 7 of 19 5 A curve is defined by the equation 3 3 3 1x y xy   . Find the values of d d y x and 2 2 d d y x at the point (2, 1). Solutions 33 13 yyx x   Differentiate wrt x, 22 d 03 (1)3 d d3 dyxy x xx yy         Sub. (2, –1) into (1), 2 2 d d3(2) 3( 1) 3 2 1 0d d d9 9 d d 1d y y x x y x y x            Differentiate (1) wrt x, 22 2 2 2 2 22 2 2 2 3 0 d d d d 6 d d d d d d 6 3 6 3 2 0 (2) d d d d d d d d d d3 6 y y yx x x x y y y yx y y x x y y yy y x x x x x x x                          Sub. (2, –1) and d d y x = −1 into (2), 2 2 2 2 2 2 2 2 2 2 d d6(2) 3( 1) 6( 1)( 1) 3 2 2( 1) 0d d d9 0d d 0d y y x x y x y x                Apply product rule

Chapter 7 Differentiation TMJC 2024 Page 8 of 19 6 2013/ACJC/II/3 (modified part) A curve C has parametric equations 2 22 , 2x t y t t t      where 0t . (a) Find d d y x in terms of t and hence find the exact value of t for which the tangent to the curve at t is parallel to the y-axis. [4] (b) Find the value of t for which the distance from the point (1, 0) to the curve is the shortest possible. [2] (c) By considering 22x and  22y , find the Cartesian equation of C. [2] Q6 Solutions (a) 2 2 2 22 d 2 21d x t t x t t t t       2 2 2 22 d 2 2 1d y t t y t t t t         2 2 2 2 2 2 d d d d d d d 2 d 2 d 2 d 2 y y t x t x y t t x t t y t x t          2 2 d 2 d 2 y t x t   When tangent parallel to the y-axis, d d y x is undefined. (Set the denominator