# Differentiate the functions given in w.r.t. $x :$ $x ^{\large\sin x} + (\sin x)^{\large\cos x}$

## 1 Answer

Toolbox:
• $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
• $\log m^{\large n}=n\log m$
• According to product rule we have
• $(uv)'=u'v+uv'$
Step 1:
Let $y=x^{\large\sin x}+(\sin x)^{\large\cos x}$
It is of the form
$y=u+v$
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Let $u=x^{\large\sin x}$
Taking $\log$ on both sides
$\log u=\log x^{\large\sin x}$
$\qquad\;=\sin x\log x$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=$$\cos x\log x+\sin x$$.\large\frac{1}{x}$
We know that $(uv)'=u'v+uv'$
$\Rightarrow \cos x\log x+\large\frac{\sin x}{x}$
$\large\frac{du}{dx}=$$u.\cos x\log x$$.\large\frac{\sin x}{x}$
Substitute the value of $u$
$\Rightarrow x^{\large\sin x}(\cos x\log x+\large\frac{\sin x}{x})$
Step 3:
Consider $v=(\sin x)^{\large\cos x}$
Taking $\log$ on both sides
$\log v=\log(\sin x)^{\large\cos x}$
$\log m^{\large n}=n\log m$
$\Rightarrow \cos x\log(\sin x)$
Differentiating with respect to $x$
$\large\frac{1}{v}\frac{dv}{dx}$$=(-\sin x)\log(\sin x)+\cos x.\large\frac{d}{dx}$$\log(\sin x)$
We know that $(uv)'=u'v+uv'$
$u=\cos x$
$v=\log\sin x$
$\Rightarrow -\sin x\log(\sin x)+\cos x.\large\frac{1}{\sin x}.\frac{d}{dx}$$(\sin x) \Rightarrow -\sin x\log(\sin x)+\large\frac{\cos x}{\sin x}.$$(\cos x)$
$\Rightarrow -\sin x\log(\sin x)+\large\frac{\cos^2 x}{\sin x}.$
$\large\frac{1}{v}\frac{dv}{dx}$$=(-\sin x)\log(\sin x)+\large\frac{\cos^2x}{\sin x} \large\frac{dv}{dx}$$=v.[(-\sin x)\log(\sin x)+\large\frac{\cos^2x}{\sin x}]$

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