How do you differentiate #y=sin^-1(3x^5+1)^3#?

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Answer 1 · 2017-04-13T13:46:19

#y'=(45x^4(3x^5+1)^2)/sqrt(1-(3x^5+1)^6)#

#y=sin^(-1)(3x^5+1)^3#

By #(sin^(-1)u)'=1/sqrt(1-u^2)# and Chain Rule,

#y'=1/sqrt(1-[(3x^5+1)^3]^2) cdot [(3x^5+1)^3]' #

By Power Rule and Cahin Rule,

#y'=1/sqrt(1-(3x^5+1)^6)cdot3(3x^5+1)^2(15x^4)#

By cleaning up a bit,

#y'=(45x^4(3x^5+1)^2)/sqrt(1-(3x^5+1)^6)#

I hope that this was clear.