How do you find the derivative of #arcsin(3-x^2)#?

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Answer 1 · 2016-12-25T06:31:21

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

We need

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

Let #y=arcsin(3-x^2)#

Then

#siny=3-x^2#

#(siny)'=(3-x^2)'#

#cosydy/dx=-2x#

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

But,

#cos^2y=1-sin^2y=1-(3-x^2)#

#cosy=sqrt(1-(3-x^2))#

#:. dy/dx=(-2x)/sqrt(1-(3-x^2))#