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,

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

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