# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$(\log x)^{\cos x}$

Solution:

Let $y=(\log x)^{\cos x}$

Taking log both the sides:

$\Rightarrow \log y=\log (\log x)^{\cos x}$

$\Rightarrow \log y=\cos x \log \log x\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$

$\Rightarrow \frac{d(\log y)}{d x}=\frac{d(\cos x \log \log x)}{d x}$

$\Rightarrow \frac{\mathrm{d}(\log y)}{d x}=\cos x \times \frac{d(\log \log x)}{d x}+\log \log x \times \frac{d(\cos x)}{d x}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\cos x \times \frac{1}{\log x} \frac{d(\log x)}{d x}+\log \log x(-\sin x)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} \& \frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\cos x}{\log x} \times \frac{1}{x}-\sin x \log \log x$

$\Rightarrow \frac{d y}{d x}=y\left\{\frac{\cos x}{x \log x}-\sin x \log \log x\right\}$

Put the value of $y=(\log x)^{\cos x}$ :

$\Rightarrow \frac{d y}{d x}=(\log x)^{\cos x}\left\{\frac{\cos x}{x \log x}-\sin x \log \log x\right\}$