Question

Differentiate the function with respect to $x$:
$(sin\,x)^x+sin^{-1}\sqrt{x}$

Answer 1

The correct answer is $\frac{dy}{dx}=(sin\,x)^x(xcot\,x+log\,sin\,x)+\frac{1}{2\sqrt{x-x^2}}$
Let $y=(sin\,x)^x+sin^{-1}\sqrt{x}$
Also,let $u=(sin\,x)^x$ and $v=sin^{-1}\sqrt{x}$
$∴y=u+v$
$⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} ....(1)$
$u=(sin\,x)^x$
$⇒log\,u=log(sin\,x)^x$
$⇒log\,u=xlog(sin\,x)$
Differentiating both sides with respect to $x$,we obtain
$⇒\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x)\times log(sin\,x)+x\times \frac{d}{dx}[log(sin\,x)]$
$⇒\frac{du}{dx}=u[1.log(sin\,x)+x.\frac{1}{sin\,x}.\frac{d}{dx}(sin\,x)]$
$⇒\frac{du}{dx}=(sin\,x)^x[log(sin\,x)+\frac{x}{sin\,x}.cosx]$
$⇒\frac{du}{dx}=(sin\,x)^x(xcot\,x+log\,sin\,x) ..(2)$
$v=sin^{-1}\sqrt{x}$
Differentiating both sides with respect to $x$, we obtain
$\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x})^2}}.\frac{d}{dx}(\sqrt{x})$
$⇒\frac{dv}{dx}=\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}$
$⇒\frac{dv}{dx}=\frac{1}{2\sqrt{x-x^2}} ....(3)$
Therefore,from (1),(2),and(3),we obtain
$\frac{dy}{dx}=(sin\,x)^x(xcot\,x+log\,sin\,x)+\frac{1}{2\sqrt{x-x^2}}$