Find dy dx y=x^ x +x^1 x - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$

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Answer

Here,
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
$=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{x}^\frac{1}{\text{x}}}$
$\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\big(\frac{1}{\text{x}}\log\text{x}\big)}$
$\big[\text{Since, e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\frac{1}{\text{x}}\log\text{x}}\Big)$
$=\text{e}^{\text{x}\log\text{x}}+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\frac{1}{\text{x}}\log\text{x}}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\log\text{x}\Big)$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{x}^\frac{1}{\text{x}}}\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +\text{x}^\frac{1}{\text{x}}\Big[\Big(\frac{1}{\text{x}}\Big)\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\Big(\frac{1}{\text{x}^2}-\frac{1}{\text{x}^2}\log\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\frac{(1-\log\text{x})}{\text{x}^2}$

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