Question
Differentiate the following functions with respect to x:
$\text{x}^{\frac{1}{\text{x}}}$
$\text{x}^{\frac{1}{\text{x}}}$
✓
Answer
Let $\text{y}=\text{x}^{\frac{1}{\text{x}}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{x}^{\frac{1}{\text{x}}}$
$\Rightarrow\log\text{y}=\frac{1}{\text{x}}\log\text{x}\ \big[\because\log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
[Using product rule]
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\times\frac{1}{\text{x}}(\log\text{x})\times\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2}-\frac{\log\text{x}}{\text{x}^2}$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{(1-\log\text{x})}{\text{x}^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{x}^\frac{1}{\text{x}}\Big[\frac{1-\log\text{x}}{\text{x}}\Big]$
Taking log on both sides,
$\log\text{y}=\log\text{x}^{\frac{1}{\text{x}}}$
$\Rightarrow\log\text{y}=\frac{1}{\text{x}}\log\text{x}\ \big[\because\log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
[Using product rule]
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\times\frac{1}{\text{x}}(\log\text{x})\times\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2}-\frac{\log\text{x}}{\text{x}^2}$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{(1-\log\text{x})}{\text{x}^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{x}^\frac{1}{\text{x}}\Big[\frac{1-\log\text{x}}{\text{x}}\Big]$
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