Differentiate the following functions with respect to x :

Solution:

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

Taking log both the sides:

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

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

Differentiating with respect to $\mathrm{x}$ :

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

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

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

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log x \times \frac{1}{\log x} \frac{d(\log x)}{d x}+\log \log x\left(\frac{1}{x} \frac{d x}{d x}\right)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\log x}{\log x}\left(\frac{1}{x} \frac{d x}{d x}\right)+\frac{\log (\log x)}{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\left\{\frac{1}{\mathrm{x}}+\frac{\log (\log \mathrm{x})}{\mathrm{x}}\right\}$

$\Rightarrow \frac{d y}{d x}=y\left\{\frac{1+\log (\log x)}{x}\right\}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(\log \mathrm{x})^{\log \mathrm{x}}\left\{\frac{1+\log (\log \mathrm{x})}{\mathrm{x}}\right\}$