Answer
$f'(x)=-\frac{x\arcsin x}{\sqrt{1-x^2}}+1$
Work Step by Step
Given: $f(x)=\sqrt{1-x^2}\arcsin x$
Let $u=\sqrt{1-x^2}$ and $v=\arcsin x$.
Find $u'$ using the chain rule:
$u'=\frac{d}{dx}(\sqrt{1-x^2})=\frac{d}{dx}((1-x^2)^{1/2})=
\frac{1}{2}(1-x^2)^{-1/2}\frac{d}{dx}(1-x^2)=\frac{1}{2\sqrt{1-x^2}}(-2x)=\frac{-x}{\sqrt{1-x^2}}$
Find $v'$:
$v'=\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}$
Using the product rule,
$f'(x)=\frac{d}{dx}(u\cdot v)$
$=u'v+uv'$
$=\frac{-x}{\sqrt{1-x^2}}\cdot \arcsin x+\sqrt{1-x^2}\cdot \frac{1}{\sqrt{1-x^2}}$
$=-\frac{x\arcsin x}{\sqrt{1-x^2}}+1$
