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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

Let $\text{u}=(\log1+\text{x})^\text{x}$
Taking log on both sides,
$\log\text{u}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\big\{\log(\log\text{x})\big\}+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\Big(\frac{1}{\log\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\text{x}}{\log\text{x}}\big(\frac{1}{\text{x}}\big)+\log\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]\ .....(\text{i})$
Again, let $\text{v}=\log\text{x}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}}\ .....(\text{ii})$
Dividing equation (i) by (ii), we get
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1+\log\text{x}(\log\log\text{x})}{\log\text{x}}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\text{x}(\log\text{x})^{\text{x}{-1}}(1+\log\text{x}\times\log\log\text{x})$

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