How do you find the derivative of #arcsin(x^2)#?

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Answer 1 · 2017-12-14T18:11:25

#dy/dx = (2x)/sqrt(1 - x^4)#

Letting #y = arcsin(x^2)#, then #siny = x^2#.

Then differentiating both sides with respect to #x#.

#cosy(dy/dx) = 2x#

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

We know that #cos^2y + sin^2y = 1#, thus #cosy = sqrt(1 - sin^2y)#.

#dy/dx = (2x)/sqrt(1 - sin^2y)#

Recall that #siny = x^2#, therefore, #sin^2y = x^4#.

#dy/dx = (2x)/sqrt(1 - x^4)#

Hopefully this helps!