Question
Differentiate the following functions with respect to x:
$(\text{x}^\text{x})\sqrt{\text{x}}$

Answer

Let $\text{y}=(\text{x}^\text{x})\sqrt{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\text{x}^\text{x}\sqrt{\text{x}})$
$\log\text{y}=\text{x}\log\text{x}+\frac{1}{2}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)+\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}+\frac{1}{2\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\sqrt{\text{x}}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}+\frac{1}{2}}\Big[\Big(\frac{2\text{x}+1}{2\text{x}}\Big)+\log\text{x}\Big]$

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