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

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Answer 1 · 2016-12-24T10:16:13

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

We need

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

#(sinx)'=cosx#

and #(x^n)'=nx^(n-1)#

So,

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

#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)^2#

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

So,

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