Differentiation Practice Answers

1 Differentiation – Practice Questions 1a Answers 1. Differentiate the following with respect to x: (a) 2𝑥! + 5𝑥 + 7 " "# 2𝑥! + 5𝑥 + 7 = 4𝑥 + 5 (b) $ ! 𝑥% + & '#! + 𝑥 " "# ) $ ! 𝑥% + & '#! + 𝑥* = " "# ) $ ! 𝑥% + & ' 𝑥(% + 𝑥* = 4 ) $ ! 𝑥'* + (−4) ) & ' 𝑥(&* + 1 = 14𝑥' − !) ' 𝑥(& + 1 = 14𝑥' − !) '#" + 1 (c) 4𝑥! + ! '√# + 7 " "# )4𝑥! + ! '√# + 7* = " "# (4𝑥! + ! ' 𝑥(# $ + 7) = 8𝑥 + )− + !* ) ! ' 𝑥(% $* = 8𝑥 − + '# % $ = 8𝑥 − + '√#% (d) 5𝑎𝑥! + 3𝑏𝑥' + 5 " "# (5𝑎𝑥! + 3𝑏𝑥' + 5) = 2(5𝑎𝑥) + 3(3𝑏𝑥!) = 10𝑎𝑥 + 9𝑏𝑥! (e) '#$(%#% #% " "# ) '#$(%#% #% * = " "# ) ' # − 4* = " "# (3𝑥(+ − 4) = − ' #$

2 (f) '√#(% √# " "# ) '√#(% √# * = " "# )3 − % √#* = " "# )3 − 4𝑥(# $* = − + ! )−4𝑥(% $* = 2𝑥(% $ = ! √#% (g) √𝑥 + + √# " "# )√𝑥 + + √#* = " "# )𝑥 # $ + 𝑥(# $* = + ! 𝑥(# $ + )− + !* 𝑥(% $ = + ! 𝑥(% $(𝑥 − 1) or = + !√# − + !√#% = #(+ !√#% 2. At x-axis, y = 0 0 = &#(% #$ 5𝑥 − 4 = 0 𝑥 = % & 𝑦 = &#(% #$ = & # − % #$ = 5𝑥(+ − 4𝑥(! ", "# = −5𝑥(! + 8𝑥(' = − & #$ + - #% At 𝑥 = % & , ", "# = − & .! "/ $ + - .! "/ % = 7 +' +0 = +!& +0 3. 𝑦 = ! "! + # " = 𝑎𝑥$% + 𝑏𝑥$& ", "# = −2𝑎𝑥(' − 𝑏𝑥(! = − !1 #% − 2 #$ At (−1, 5), ", "# = 4 = − !1 ((+)% − 2 ((+)$ 4 = − !1 (+ − 2 + 4 = 2𝑎 − 𝑏 ----- (1)

3 At (−1, 5), 5 = ! ($&)! + # $& 5 = 𝑎 − 𝑏 𝑎 = 5 + 𝑏 ----- (2) 𝑆𝑢𝑏 (2) 𝑖𝑛𝑡𝑜 (1) 4 = 2(5 + 𝑏) − 𝑏 4 = 10 + 2𝑏 − 𝑏 4 − 10 = 𝑏 𝑏 = −6 𝑎 = 5 − 6 = −1 4. CHAIN RULE (a) (2𝑥 + 5)$ " "# [(2𝑥 + 5)$] = 7(2𝑥 + 5)0(2) = 14(2𝑥 + 5)0 (b) 3(𝑥 + 4)& " "# [3(𝑥 + 4)&] = 3(5)(𝑥 + 4)%(1) = 15(𝑥 + 4)% (c) ! ' ) # 0 − 1* % " "# < ! ' ) # 0 − 1* % = = ! ' (4) ) # 0 − 1* ' ) + 0* = % 5 ) # 0 − 1* ' (d) + '#6! " "# ) + '#6!* = " "# [(3𝑥 + 2)(+] = −(3𝑥 + 2)(!(3) = − ' ('#6!)$ (e) +! !6'#$ " "# ) +! !6'#$* = 12 " "# (2 + 3𝑥!)(+

www.ApexEducators.com 4 = 12(−1)(2 + 3𝑥!)(!(6𝑥) = − $!# (!6'#$)$ (f) √5𝑥! + 6 7 "# >√5𝑥! + 6? = " "# (5𝑥! + 6) # $ = + ! (5𝑥! + 6)(# $(10𝑥) = &# √&#$60 5. 𝑦 = (3𝑥! − 5𝑥 + 3)' ", "# = 3(3𝑥! − 5𝑥 + 3)!(6𝑥 − 5) At 𝑥 = 1, ", "# = 3(3(1)! − 5(1) + 3)!(6(1) − 5) = 3(3 − 5 + 3)!(6 − 5) = 3(1)!(1) = 3 6. 𝑦 = 2 + +! ('#(%)$ = 2 + 12(3𝑥 − 4)(! ", "# = 12(−2)(3𝑥 − 4)('(3) = − $! ('#(%)%. At (2, 5), ", "# = − $! ('(!)(%)% = − $! - = −9 Thus 𝑦 − 5 = −9(𝑥 − 2) 𝑦 = −9𝑥 + 18 + 5 𝑦 = −9𝑥 + 23 7. 𝑦 = 1 !62# = 𝑎(2 + 𝑏𝑥)(+ ", "# = 𝑎(−1)(2 + 𝑏𝑥)(!(𝑏) = − 12 (!62#)$ At x = 1, ", "# = ' & = − 12 (!62)$ 3(2 + 𝑏)! = −5𝑎𝑏 ------ (1) At x = 1, y = 1, 1 = 1 !62

www.ApexEducators.com 5 𝑎 = 2 + 𝑏 ------ (2) 𝑆𝑢𝑏 (2) 𝑖𝑛𝑡𝑜 (1) 3(2 + 𝑏)! = −5𝑏(2 + 𝑏) 3(4 + 4𝑏 + 𝑏!) = −10𝑏 − 5𝑏! 12 + 12𝑏 + 3𝑏! = −10𝑏 − 5𝑏! 8𝑏! + 22𝑏 + 12 = 0 (𝑏 + 2)(8𝑏 + 6) = 0 𝑏 = −2 or 𝑏 = − ' % 𝑎 = 2 − 2 = 0 or 𝑎 = 2 − ' % = 1 + % Reject 𝑎 =0 and 𝑏 = −2

www.ApexEducators.com 6 Differentiation – Practice Questions 1b Answers 1. Answer (a) (3𝑥+2)(2−𝑥!) ""#[(3𝑥+2)(2−𝑥!)]=(3𝑥+2)""#(2−𝑥!)+(2−𝑥!)""#(3𝑥+2) =(3𝑥+2)(−2𝑥)+(2−𝑥!)(3) =−6𝑥!−4𝑥+6−3𝑥! =−9𝑥!−4𝑥+6 (b) (𝑥+1)'(𝑥+3)& ""#[(𝑥+1)'(𝑥+3)&]= (𝑥+1)'""#(𝑥+3)&+(𝑥+3)&""#(𝑥+1)' =(𝑥+1)'(5)(𝑥+3)%+(𝑥+3)&(3)(𝑥+1)! =(𝑥+1)!(𝑥+3)%[5(𝑥+1)+(𝑥+3)(3)] =(𝑥+1)!(𝑥+3)%[5𝑥+5+3𝑥+9] =(𝑥+1)!(𝑥+3)%(8𝑥+14) =2(𝑥+1)!(𝑥+3)%(4𝑥+7) (c) (𝑥+5)'(𝑥−4)0 ""#[(𝑥+5)'(𝑥−4)0]= (𝑥+5)'""#(𝑥−4)0+(𝑥−4)0""#(𝑥+5)' =(𝑥+5)'(6)(𝑥−4)&+(𝑥−4)0(3)(𝑥+5)! =(𝑥+5)!(𝑥−4)&[6(𝑥+5)+3(𝑥−4)] =(𝑥+5)!(𝑥−4)&[6𝑥+30+3𝑥−12] =(𝑥+5)!(𝑥−4)&(9𝑥+18) =9(𝑥+5)!(𝑥−4)&(𝑥+2) (d) (3𝑥−1)√2𝑥!+3=(3𝑥−1)(2𝑥!+3)#$ ""#G(3𝑥−1)(2𝑥!+3)#$H= (3𝑥−1)""#(2𝑥!+3)#$+(2𝑥!+3)#$""#(3𝑥−1) =(3𝑥−1))+!*(2𝑥!+3)(#$(4𝑥)+(2𝑥!+3)#$(3) =(2𝑥!+3)(#$[(3𝑥−1)(2𝑥)+(2𝑥!+3)+(3)] =(2𝑥!+3)(#$[6𝑥!−2𝑥+6𝑥!+9]

www.ApexEducators.com 7 =80#$(!#60#$659(!#$6')#$ =+!#$(!#65√!#$6' (e) (𝑥'+𝑥!)(𝑥−2)$ ""#[(𝑥'+𝑥!)(𝑥−2)$]=(𝑥'+𝑥!)""#(𝑥−2)$+(𝑥−2)$""#(𝑥'+𝑥!) =(𝑥'+𝑥!)(7)(𝑥−2)0+(𝑥−2)$(3𝑥!+2𝑥) =(𝑥−2)0[7(𝑥'+𝑥!)+(𝑥−2)(3𝑥!+2𝑥)] =(𝑥−2)0[7𝑥'+7𝑥!+3𝑥'+2𝑥!−6𝑥!−4𝑥] =(𝑥−2)0[10𝑥'+3𝑥!−4𝑥] =𝑥(𝑥−2)0(10𝑥!+3𝑥−4) =𝑥(𝑥−2)0(2𝑥−1)(5𝑥+4) 2. 𝑦=(𝑥+3)%(𝑥−5)$ ","#=(𝑥+3)%(7)(𝑥−5)0+(𝑥−5)$(4)(𝑥+3)' =(𝑥+3)'(𝑥−5)0[7(𝑥+3)+(4)(𝑥−5)] =(𝑥+3)'(𝑥−5)0[7𝑥+21+4𝑥−20] =(𝑥+3)'(𝑥−5)0(11𝑥+1) When ","#=0, (𝑥+3)'(𝑥−5)0(11𝑥+1)=0 Thus (𝑥+3)'=0, (𝑥−5)0=0, (11𝑥+1)=0 𝑥=−3,5,−+++ 3. 𝑦=(3𝑥+2)(2−𝑥)(+ ","#=(3𝑥+2)(−1)(2−𝑥)(!(−1)+ (2−𝑥)(+(3) =(2−𝑥)(![(3𝑥+2)+(2−𝑥)+(3)] =(2−𝑥)(![3𝑥+2+6−3𝑥] =-(!(#)$ When ","#=8, 8= -(!(#)$

www.ApexEducators.com 8 (2−𝑥)!=1 2−𝑥=±1 𝑥=1, 3 4. Answer (a) '#6!+(%# ""#)'#6!+(%#*=(+(%#)(')(('#6!)((%)(+(%#)$ ='(+!#6+!#6-(+(%#)$ =++(+(%#)$ (b) #$!#(+ ""#)#$!#(+*=(!#(+)(!#)((#$)(!)(!#(+)$ =%#$(!#(!#$(!#(+)$ =!#$(!#(!#(+)$ =!#(#(+)(!#(+)$ (c) √#'6# ""#)√#'6#*=('6#).#$/:#&#$;(##$(+)('6#)$ =#$#&#$[('6#)(!#]('6#)$ =['6#(!#]!√#('6#)$ ='(#!√#('6#)$ (d) !#($√#6+ ""#)!#($√#6+*=(#6+)#$(!)((!#($).#$/(#6+)&#$ #6+ =#$(#6+)&#$[!(!)((!#($)] #6+ =[%(!#6$] !?(#6+)% =++(!# !?(#6+)%

www.ApexEducators.com 9 8. 𝑦=#6!√'#6+ ","#=('#6+)#$(+)((#6!).#$/('#6+)&#$(') '#6+ =#$ ('#6+)&#$[!('#6+)((#6!)(')] '#6+ =[!('#6+)((#6!)(')] !('#6+)%$ =0#6!('#(0 !('#6+)%$ ='#(% !('#6+)%$ (a) At 𝑥=1, ","#='(+)(% !('6+)%$ =−++0 (b) 𝑦=+6!√'6+='! 𝑦−'!=−++0(𝑥−1) 𝑦=−++0𝑥+!&+0 16𝑦=−𝑥+25 9. Answer (a) 2𝑥+9𝑦=3 ------ (1) 𝑥𝑦+𝑦+2=0 ------ (2) (1): 𝑥='(5,! ------ (3) Sub (3) into (2) )'(5,!*𝑦+𝑦+2=0 (3−9𝑦)𝑦+2𝑦+4=0 3𝑦−9𝑦!+2𝑦+4=0 −9𝑦!+5𝑦+4=0 (𝑦−1)(−9𝑦−4)=0 𝑦=1 or 𝑦=−%5 𝑥='(5! or 𝑥='(5.(!'/! 𝑥=−3 or 𝑥=$!

www.ApexEducators.com 10 (b) Answer 𝑥𝑦+𝑦+2=0 𝑥𝑦+𝑦=−2 𝑦(𝑥+1)=−2 𝑦=−!(#6+)=−2(𝑥+1)(+ ","#=−2(−1)(𝑥+1)(!=!(#6+)$ or ","#=−(#6+)())(!(+)(#6+)$=!(#6+)$ At 𝑥=−3, ","#=!(('6+)$=+! or 𝑥=$! , ","#=!('.&6+)$=--+