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,

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

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

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

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