Question

Differentiate the function w.r.t. $x$.
$(\sin x{)}^x+{\sin }^{-1}\sqrt{x}$

Answer 1

Let $y=(\sin x{)}^x+{\sin }^{-1}\sqrt{x}$
Also, let $u=(\sin x{)}^x$ and $v={\sin }^{-1}\sqrt{x}$
$\therefore y=u+v$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ...(1)
$u=(\sin x{)}^x$
$\Rightarrow \log u=\log (\sin x{)}^x$
$\Rightarrow \log u=x\log \left(\sin x\right)$
Differentiating both sides with respect to $x$, we obtain
$\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(x\right)\times \log \left(\sin x\right)+x\times \frac{d}{dx}\left[\log \right(\sin x\left)\right]$
$\Rightarrow \frac{du}{dx}=u[1.\log (\sin x)+x.\frac{1}{\sin x}.\frac{d}{dx}(\sin x\left)\right]$
$\Rightarrow \frac{du}{dx}=(\sin x{)}^x\left[\log \right(\sin x)+\frac{x}{\sin x}.\cos x]$
$\Rightarrow \frac{du}{dx}=\left(\sin x{)}^x\right(x\cot x+\log \sin x)$ ...(2)
$v={\sin }^{-1}\sqrt{x}$
Differentiating both sides with respect to $x$, we obtain
$\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}.\frac{d}{dx}\left(\sqrt{x}\right)$
$\Rightarrow \frac{dv}{dx}=\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}$
$\Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{x-x^2}}$ ...(3)
Therefore, from (1), (2) and (3), we obtain
$\frac{dy}{dx}=\left(\sin x{)}^x\right(x\cot x+\log \sin x)+\frac{1}{2\sqrt{x-x^2}}$