If #y=arcsin(4x^2)#, then what is #dy/dx#?

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Answer 1 · 2016-12-19T14:19:51

The answer is #=(8x)/sqrt(1-16x^4)#

By definition

If #y=arcsinx#, then, #siny=x#

Here, we have

#y=arcsin4x^2#

So,

#siny=4x^2#

Differentiating

#cosydy/dx=8x#

#dy/dx=(8x)/cosy#

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

#cos^2y=1-sin^2y=1-(4x^2)^2#

#=1-16x^4#

#cosy=sqrt(1-16x^4)#

#dy/dx=(8x)/sqrt(1-16x^4)#