How do you differentiate #f(x)=sin^-1(x^3)#?

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Answer 1 · 2017-02-05T09:01:07

The answer is #=(3x^2)/(sqrt(1-x^6))#

Let #y=sin^-1x^3#

So,

#siny=x^3#

Differentiating wrt #x#

#cosy(dy/dx)=3x^2#

#dy/dx=(3x^2)/(cosy)#

But,

#sin^2y+cos^2y=1#

#cos^2y=1-sin^2y=1-x^6#

#cosy=sqrt(1-x^6)#

Therefore,

#dy/dx=f'(x)=(3x^2)/(sqrt(1-x^6))#