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