ACJC 2025 Differentiation and Applications Summary

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Differentiation and Applications / Summary / Page 1 of 7 SUMMARY: Differentiation and Applications Standard Formulae Let u, v be functions of x, and a and b be constants. Basic Rules Examples Addition/subtraction: 5d6 23d xxx  + − = 2 (5x4) + 6 (−1  x−2) − 0 Product Rule: ( ) x uvx vuuvx d d d d d d += ( )( ) 2d 1 5 4d xxx  ++ = (x2 + 1)(5) + (2x)(5x + 4) Quotient Rule: 2 d d d d d d v x vux uv v u x − =     3 2 d4 d2 x xx   + = ( )( ) ( ) ( ) 2 2 3 22 2 12 4 2 2 x x x x x +− + Chain Rule: ( ) 72d1 31d2 xx  + = ( ) 627 3 1 62 xx+ Derivatives of Exponential Functions (Let a be a positive constant.) Examples ( )d eed xx x = ( ) ( ) ( ) ( )ffd e f ed xx xx = ( ) ( )sin sind e cos ed xx xx = ( )d lnd xxa a ax = ( ) ( ) ( ) ( )ffd f lnd xx a x a ax = Note: Use lny = ln(ax) with implicit differentiation if you do not want to memorise formula. sin2 xy= ( )ln sin ln 2yx= Differentiating w.r.t. x, ( )1d cos ln 2d y xyx = ( )( ) sind cos ln 2 2d xy xx= Derivatives of Logarithmic Functions Examples ( )d1 lnd xxx = ( )( ) ( ) ( ) fd ln fdf xxxx = ( )( )d 1 d 1 1ln ln 1 ln( 1)d 1 d 1 1 x xxx x x x x  +  = + − − = − − + − ( )d1 log log ed aaxxx = ( )( ) ( ) ( ) fd log f log edf aa xxxx = Note: Change base from loga to ln if you do not want to memorise formula. ( )( ) 2 2 5 2 d d ln(1 ) 1 2log 1d d ln 5 ln 5 1 xxxx x x  + + = =   +

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Differentiation and Applications / Summary / Page 2 of 7 Derivatives of Trigonometric Functions (Note: x is in radians) ( )d sin cosd xxx = ( ) xxxx tansecsecd d = (MF27) ( )d cos sind xxx =− ( )d cosec cosec cotd x x xx =− ( ) 2d tan secd xxx = ( ) 2d cot cosecd xxx =− Derivatives of Trigonometric Functions using Chain Rule Examples ( ) ( ) 1d sin sin cosd nnx n x xx −= ( ) ( ) 32d sin 3 sin cosd x x xx = ( ) ( ) 12d tan tan secd nnx n x xx −= ( ) ( ) 22d tan 2 tan secd x x xx = ( ) ( ) 1d sec sec sec tand nnx n x x xx −= ( ) ( ) 22d sec 2 sec sec tan 2sec tand x x x x x xx == Differentiation Formula in MF27 Derivatives f ( )x f ( )x 1sin x− 2 1 1 x− 1cos x− 2 1 1 x − − 1tan x− 2 1 1 x+ secx sec tanxx Derivatives of Inverse Trigonometric Functions Examples ( ) ( ) ( ) 1 2 fd sin fd 1f xxx x −  = − ( ) 1 2 d2 sin 2d 14 xx x − = − ( ) ( ) ( ) 1 2 fd cos fd 1f xxx x − − = − ( ) 2 1 2 d seccos tand 1 tan xxx x − − = − ( ) ( ) ( ) 1 2 fd tan fd 1f xxx x −  = + ( ) 1 2 de tan ed 1 e x x xx − = +

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Differentiation and Applications / Summary / Page 3 of 7 Parametric Differentiation Example Use d d d .d d d yy xx  = (or d dd ddd y yx x = ) sin 2xt= , cos 2yt= d 2cos 2d x tt = , d 2sin 2d y tt =− d 2sin 2 tan 2d 2cos 2 yt txt − = =− Implicit Differentiation Example ( ) 2dd 2dd yyyxx = ( ) ( ) 22dd 2dd yx y x y xxx =+ Given 22 2x xy y+ + = , find d d y x . 22 2x xy y+ + = Differentiating w.r.t. x, dd2 2 0 dd yyx x y y xx + + + = ( ) d22 d yy x y x x + =− − d ( 2 ) d2 y y x x y x −+= +

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Differentiation and Applications / Summary / Page 4 of 7 Graph of ( )fyx = Graph of ( )fyx= Graph of ( )fyx = Mathematical Feature Geometrical Feature 1 Stationary points i.e., d 0d y x= at x = a Max, Min or stationary point of inflexion at x = a Intercept at x = a 2(a) Vertical asymptote at xp= Vertical asymptote at x = p 2(b) Horizontal asymptote at yq= Horizontal asymptote at y = 0 2(c) Oblique asymptote at y mx c=+ Horizontal asymptote at y = m Start sketching the graph of f ( )yx = from left to right, using the following checks: 3(a) d 0d y x in interval (a,b) Curve increasing in interval (a,b) Curve above x-axis (b) d 0d y x in interval (a,b) Curve decreasing in interval (a,b) Curve below x-axis 4(a) 2 2 d 0d y x  in interval (a,b) Curve is concave upwards in interval (a,b) Curve is increasing in interval (a,b) (b) 2 2 d 0d y x  in interval (a,b) Curve is concave downwards in interval (a,b) Curve is decreasing in interval (a,b) 5 2 2 d 0d y x = at x = k Change in curvature of curve There is a stationary point (max/min/inflexion) at x = k Example The diagram below shows the graph of 1x= . The graph has a horizontal asymptote at 1y=− , a vertical asymptote at 1x= , a maximum point at 3x= , and a point of inflexion at 4x= . x y

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Differentiation and Applications / Summary / Page 5 of 7 Applications of Differentiation (A) Tangents & Normals • The equation of the tangent to a curve y = f(x) at 11( , )P x y is ( )11y y m x x− = − , where m is the value of x y d d when 1xx= . • The equation of the normal to a curve y = f(x) at 11( , )P x y is )(1 11 xxmyy −−=− , where m is the value of x y d d when 1xx= . • Consider a curve C with the following parametric equations: 1 sinxt=+ , cosyt= 0 πt . The following are equivalent statements (i) Find the equation of the tangent at the point with parameter p. (ii) Find the equation of tangent at t = p. (iii) Find the equation of tangent at the point (1 sin ,cos )pp+ . Example A curve is defined by the parametric equations 32xt=+ and 3yt t=+ , where t is a non -zero parameter. Find the equation of the tangent at the point P with parameter p and hence write down the equation of tangent at the point where x = 11. Solution: d 3d x t = , 2 d3 1d y t t =− 2 d d d 3 1 1d d d 3 y y t x t x t =  = −  2 1 3 1 t−= When t = p, 2 d 1 1 d3 y x p =− , 32xp=+ , 3yp p=+ Equation of tangent at 33 2,pp p  ++ is ( )2 3 1 1 323y p x p p p − + = − − +    When x = 11, 3 2 11 3pp+ =  = Thus 334 3y= + = , and 2 d 1 1 2 d 3 3 9 y x = − = Equation of tangent when x = 11 is ( )119 24 −    =− xy 2 14 99 xy = + Step 1 Find d d y x in terms of t. Step 2 At the point P with parameter p, sub t = p to find x, y and d d y x in terms of p. Step 3 Use the formula 11 ()y y m x x− = − or y mx c=+ at the point P where t = p Important to find the specific value of p when x = 11 as this allows you to find values of y and d d y x

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Differentiation and Applications / Summary / Page 6 of 7 (B) Maxima/Minima Example A closed circular cylinder of radius x is inscribed in a right circular cone of radius 2 h and vertical height 3h, where h is a constant. One circular end of the cylinder lies on the base of the cone and the circumference of the other circular end is in contact with the inner surface of the cone. If x is made to vary, find the maximum volume of the cylinder in terms of h. [4] Solution: Let y be the height of the cylinder. 2πV x y= By similar triangles, 3 32 33 2 3 3 2 h y x hh h y x y h x − =  − =  = − 2 2 Hence π 13π ( )2 V x y x h x = =− For maximum V, 2d 3 9 3π (2 ) π(3 ) 0 (6 ) 0d 2 2 V h x x x h xx = − =  − = 40 (rejected) or 3x x h = = Check that 2 2 4dwhen , 6 π 12π 6π 03d Vx h h h h x= = − =−  . 2 234 3 4 16 16π 3 π ( ) π3 2 3 9 9V h h h h h h      = − = =             Hence V is maximum at 316 π9 h . 2h 3h x Step 1 Since V is to be maximum, we start off by expressing V in terms of 2 variables, x and y. Step 2 Identify the constraint to reduce the no. of variables to 1. Possible constraints include similar triangles,