Question

Differentiate the function w.r.t. $x$.
$x^x-2^{\sin x}$

Answer 1

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