# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\left(x^{x}\right) \sqrt{x}$

Solution:

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

Taking log both the sides:

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

$\Rightarrow \log y=\log (x)^{x}+\log \sqrt{x}\{\log (a b)=\log a+\log b\}$

$\Rightarrow \log y=\log (x)^{x}+\log x^{\frac{1}{2}}$

$\Rightarrow \log y=x \log x+\frac{1}{2} \log x$

$\left\{\log x^{a}=a \log x\right\}$

$\Rightarrow \log y=\left(x+\frac{1}{2}\right) \log x$

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

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\left(\mathrm{x}+\frac{1}{2}\right) \log \mathrm{x}\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\left(\mathrm{x}+\frac{1}{2}\right) \times \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}+\log \mathrm{x} \times \frac{\mathrm{d}\left(\mathrm{x}+\frac{1}{2}\right)}{\mathrm{dx}}$

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

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\left(x+\frac{1}{2}\right) \times \frac{1}{x} \frac{d x}{d x}+\log x \frac{d x}{d x}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{(2 x+1)}{2} \times \frac{1}{x}+\log x$

$\Rightarrow \frac{d y}{d x}=y\left\{\frac{(2 x+1)}{2 x}+\log x\right\}$

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

$\Rightarrow \frac{d y}{d x}=(x)^{x} \sqrt{x}\left\{\frac{(2 x+1)}{2 x}+\log x\right\}$

$\Rightarrow \frac{d y}{d x}=(x)^{x} \sqrt{x}\left\{\frac{2 x}{2 x}+\frac{1}{2 x}+\log x\right\}$

$\Rightarrow \frac{d y}{d x}=(x)^{x} \sqrt{x}\left\{1+\frac{1}{2 x}+\log x\right\}$