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Answer 1 · 2017-04-21T15:00:39

#f'(x)=4arcsin(x)1/(sqrt(1-x^2))#

#f'(x)=4arcsin(x) (arcsin(x))'#

To calculate #(arcsin(x))'# use the inverse rule:

#y=f^{-1}(x)\Rightarrow y'=1/(f'(y))#

Hence if #y=arcsin(x)#, we have #x=sin(y), x'=cos(y)#, hence #y'=1/cos(y)=1/cos(arcsin(x))#.

In order to compute #cos(arcsin(x))#, consider

#cos^2(arcsin(x))+sin^2(arcsin(x))=1#, but #sin^2(arcsin(x))=x^2#, hence

#cos(arcsin(x))=sqrt(1-x^2)#. Hence we get the answer