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