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

Answer

Given: xx - 2sin x
Let y = xx - 2sin x
Let y = u - v
$\Rightarrow$ u = xx and v = 2sin x
For, u = xx
Taking log on both sides, we get
log u = log xx
$\Rightarrow$ log u = x.log(x)
Now, differentiate both sides with respect to x
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\mathrm{x})]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\mathrm{x}}+\log \mathrm{x} \cdot(1)\right]$ 
$\Rightarrow \frac{d u}{d x}=x^{x}(1+\log x)$
For, v = 2sin x
Taking log on both sides, we get
log v = log 2sin x
$\Rightarrow$ log v = sin x.log (2)
Now, differentiate both sides with respect to x
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{v})=\frac{\mathrm{d}}{\mathrm{dx}}[\sin \mathrm{x} \cdot \log (2)]$ 
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\log 2 \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})$ 
$\Rightarrow \frac{d v}{d x}=v[\log 2 .(\cos x)]$ 
$\Rightarrow \frac{d v}{d x}=2^{\sin x} \cdot \cos x \cdot \log 2$ 
Because, y = u - v 
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$ 
$\frac{dy}{dx}$ = xx(1 + log x) - 2sin x.cosx.log 2

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