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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$(\log\text{x})^\text{x}$
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Answer

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

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