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.

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