## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dy}{dx} = \ sin^{-1}{x}$
Given that $y$ = $x\sin^{-1}{{x}}+\sqrt{(1-x^2)}$, on applying the product rule and the chain rule of the derivative, we get: $\frac{dy}{dx}$ = ${{1\times\sin^{-1}{x}+x\times{\frac{1}{\sqrt{(1-x^2)}}}}+{\frac{1}{2\sqrt{(1-x^2)}}}}\frac{d{(1-x^2)}}{dx}$ or $\frac{dy}{dx}$ = $\ sin^{-1}{x}+{ \frac {x}{\sqrt{({1-x^2})}}-{ \frac {x}{\sqrt{({1-x^2})}}}} =\ sin^{-1}{x}$ The final answer is:$\frac{dy}{dx} = \ sin^{-1}{x}$