SAJC Chapter 5 Techniques of Differentiation (Tutor) 2022

SAJC 2022 JC 2 H1 Mathematics Page 1 of 20 Chapter 5 (Pure Mathematics) Techniques of Differentiation Objectives At the end of the chapter, you should be able to: (a) understand the derivative of ( )f x as the gradient of the tangent to the graph of ( )fyx= at a point; (b) use of standard notations ( )f ' x and d d y x ; (c) differentiate nx for any rational n, xe , ln x , together with constant multiples, sums and differences; (d) use the chain rule to differentiate a composition of 2 or 3 functions; (e) find the approximate value of a derivative at a given point using a graphing calculator Content 5.1 Introduction 5.2 Derivatives of Basic Functions 5.3 Rules of Differentiation 5.3.1 Extension of Derivatives of Basic Functions 5.4 Finding Numerical Values of Derivatives for Given Values of x by GC 5.5 Higher Order Derivatives References New Syllabus Additional Mathematics (8th Edition), Shinglee Publishers Pte Ltd. Relevant Resources • http://mathinsite.bmth.ac.uk/applet/diffchords/diffchords.html (Applet to explore graphically the differentiation of a quadratic curve of 2yx= • http://www.slu.edu/classes/maymk/GeoGebra/TangentToDerivatives.html (Applet to explore differentiation of being the slope of the line tangent to a curve)

SAJC 2022 JC 2 H1 Mathematics Page 2 of 20 5.1 Introduction View the following video 1. https://www.youtube.com/watch?time_continue=15&v=EKvHQc3QEow or 2. Scan the QR code on the right. You should see this page: Note: 1. The process of finding the derivative of a function is called differentiation. d d y x means we differentiate the function y with respect to x. 2. If ( )fyx= , we may write ( )dd f f '( )dd y xxxx== ( )f' x is known as the first derivative of ( )f x . 3. For a curve ( )fyx= , d d y x denotes gradient of the tangent to the curve ( )fyx= at the point ( )( ),fxx . 4. d d y x also denotes the rate of change of the function y with respect to x. Alternatively, you can use the QR code:

SAJC 2022 JC 2 H1 Mathematics Page 3 of 20 5.2 Derivatives of Basic Functions Basic Functions y d d y x Algebraic (for any rational n) nx 1nnx − Exponential ex ex Logarithmic ln x 1 x Example 1 (Differentiate algebraic functions) Differentiate the following with respect to x: (a) 6xy = 6 5d 6d yx y xx = = (b) 4yx −= 4 5 5 d 4d 4 yx y xx x − − = =− =− (c) yx= 1 2 1 2 1 2 d1 d2 1 2 1 2 y x x y xx x x − == = = = (d) 6y= 6 d 0d y y x = =

SAJC 2022 JC 2 H1 Mathematics Page 4 of 20 5.3 Rules of Differentiation Suppose ( )f x and ( )g x are functions of x and a and b are constants. Rules Example 1 Constant Multiple of Function     ( ) dd f ( ) f ( )dd f' a x a xxx ax = = 2d 20d xx  = 20(2 ) 40xx= d 5d xex −= 5( ) 5xxee− =− d ln 1 1 1 d 3 3 3 x x x x    ==     2 Sum and Differences of Functions d d d( f ( ) g( )) f ( ) g( )d d d f '( ) g'( ) a x b x a x b xx x x a x b x  =  =+ 2d1 20 3ln ) 40 3d 340 x x xxx x x  + = +   =+ 3 Chain Rule d d d d d d y y u x u x= where )(f uy= and )(g xu = See Example 3 d d d d d d d d y y u v x u v x=   where )(f uy= , ( )u g v= and ( )v h x= See Example 4(e)

SAJC 2022 JC 2 H1 Mathematics Page 5 of 20 Example 2 Differentiate the following with respect to x: (a) 3 (b) 2x (c) 32x (d) 34 x (e) 2 1 4x (f) 5 xe (g) 1 ln5 x (h) 267xx+ (i) 1 37e 2 ln 5x x − −+ (j) 2ln 2 x x+ (k) 5 75xx x −+ (l) 2 3x x  − Solution: (a) ( )d 30dx = (b) ( ) ( ) ( ) dd 22dd 2 1 2 xxxx = == (c) ( ) ( ) ( ) 33 2 2 dd 22dd 23 6 xxxx x x = = = (d) ( ) 1 3 3 2 3 2 3 2 3 dd 44dd 14 3 44 33 xxxx x xx − =   =   == (e) ( ) ( ) 2 2 3 3 d 1 1 d d 4 4 d 1 24 1 2 xx x x x x − −  = =− =− (f) ( ) ( )dd 55dd 5 xx x eexx e = = (g) ( )d 1 1 dln lnd 5 5 d 11 5 1 5 xxxx x x  = =  = (h) ( ) ( ) ( ) 2d 6 7 6 2 7 1d 12 7 x x xx x + = + =+ (i) ( ) 1 3 4 3 3 4 d 7e 2 ln 5d 17 2 0 3 27 3 x x x xx ex e x − −  −+  = − − +   =+ (j) 2d ln 1 1 2d 2 2 1 22 x xxxx xx    + = +       =+ (k) 5 5 9 1 1 2 2 2 d 7 5 d d 7 5 d d 75d xx x x xx x x x x x x xx −  −+  = − + = − +  7 1 3 222 7 2 3 9 7 5 2 2 2 9 7 5 2 2 2 x x x x x x −− = − − = − − (l) ( ) 2 2 2 d3 d d9 6d 19 91 xx x xxx x x −  − = − + = + − =− Note: The use of the product rule and the quotient rule is out of the H1 Mathematics syllabus. You may use them if you wish to.

SAJC 2022 JC 2 H1 Mathematics Page 6 of 20 5.3.1 Extension of Derivatives of Basic Functions Using Chain Rule Type of Function Basic General y d d y x y d d y x Algebraic nx 1nnx − ( )f n x ( ) ( ) 1 f f ' n n x x −  Exponential xe xe ( )f x e ( ) ( ) f f' x ex Logarithmic ln x 1 x ( )ln f x ( ) ( ) f' f x x Example 3 [Chain Rule] Differentiate the following with respect to x: (a) ( ) 2368yx=+ (b) ( ) 4 lnyx= (c) 1 6x y ex = − (d) ( ) 2ln 1yx=+ Solution: (a) 32Let 6 8u x y u= +  = 32dd (6 8) 18dd u xxxx= + = and ( ) 2dd 2dd y uuuu== By Chain Rule, ( )( ) ( ) 2 32 52 d d d d d d 2 18 2(6 8) 18 216 288 y y u x u x ux xx xx = = =+ =+ Shorter working: ( ) ( ) 23 32 52 68 d 2(6 8) 18d 216 288 yx y xxx xx =+ =+ =+ (b) 4Let lnu x y u=  = d1 d u xx= and 3d 4d y uu = Chain Rule, ( ) ( ) ( ) 3 3 3 d d d d d d 14 14 ln 4 ln y y u x u x u x x x x x = =  =  = By Shorter working: ( ) ( ) ( ) 4 3 3 ln 4 lnd1 4 lnd yx xy xx x x = ==  Is this correct? ( ) 4 ln 4ln d4 d y x x y xx == =

SAJC 2022 JC 2 H1 Mathematics Page 7 of 20 (c) 1 2Let 6 xu e x y u − = −  = d 6d xu ex =− and 3 2d1 d2 y uu −=− By Chain Rule, ( ) ( )( ) 3 2 3 2 d d d d d d 1 62 1 662 x xx y y u x u x ue e e x − − = = − − =− − − Shorter working: ( ) ( ) ( ) ( )( ) 1 2 3 2 3 2 1 6 6 d1 66d2 1 662 x x xx xx y ex y e x y e x ex e e x − − − = − =− = − − − =− − − (d) 2Let ( 1) lnu x y u= +  = ( ) 2dd 12dd u xxxx= + = and ( )d d 1 lndd y uu u u== By Chain Rule, 2 d d d d d d 1 (2 ) 2 1 y y u x u x xu x x = = = + Shorter working: ( ) 2 2 2 2 Let ln( 1) d 1 d 1d ( 1) d d2 d1 yx y xx x x yx xx =+ =+ + = +

SAJC 2022 JC 2 H1 Mathematics Page 8 of 20 Example 4 Differentiate the following with respect to x: (a) 2 4 2 5 23 xyx x= − + , (b) 2 354 xxye +−= , (c) ( ) 24ln 2e xy ex −=− , (d) 4252ln 3 xy  −=  . (e) 2 7exxy −= Solution: (a) 2 4 2 1 2 4 2 5 23 5 23 xyx x x xx = − + = − + ( ) 3 2 4 3 2 3 2 4 23 d 1 5 2 2 5 2 2d 4 3 3 1 5 2 10 224 3 3 y x x xxxx xx xxx − − −  = − + − − +   = − + + +  (b) ( ) ( ) 2 2 2 35 35 35 4 d 4 2 3d 4 2 3 xx xx xx ye y exx xe +− +− +− = =+ =+ (c) ( ) ( ) 24 4 24 4 24 ln 2e 2 2 4ed d 2e 2 8e 2e x x x x x y ex exy x ex ex ex − − − − − =− −− = − += − (d) ( ) ( ) 42 2 2 52ln 3 4 ln 5 2 ln 3 4ln 5 2 4ln 3 xy x x  −=  = − − = − − 2 2 d 10 4d 5 2 40 52 yx xx x x =  − = − (e) 2 7Let e xxu y u −=  = 1 2d 1 1 d2 2 y uu u − == 2Let 7 e vv x x u= −  = d ed vu v = d 27d v xx =−

SAJC 2022 JC 2 H1 Mathematics Page 9 of 20 By Chain Rule, ( )( ) ( )( ) ( ) 2 2 2 7 7 7 d d d d d d d d 1 e 2 7 2 1 e 2 7 2e 1 2 7 e2 v xx xx xx y y u v x u v x x u x x − − − =   =− =− =− Note: For Example 4(d), if we had not simplified our expression for y before performing di