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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability5 Marks
Question
Differentiate the function $(\log x)^x + x^{\log x}$ w.r.t. $x$.
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Answer

Given: $(\log x)^x + x^{\log x}$
Let $y = (\log x)^x + x^{\log x}$
Let $y = u + v$
$\Rightarrow u = (\log x)^x$ and $v = x^{\log x}$
For, $u = (\log x)^x$
Taking \log on both sides, we get
$\log u = \log (\log x)^x$​​​​​​​
$\Rightarrow \log u = x.\log (\log (x))$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\log \mathrm{x})]$
$\left.\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log (\log \mathrm{x})\right)+\log (\log \mathrm{x}) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\log \mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log (\log \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=(\log \mathrm{x})^{\mathrm{x}}\left[\frac{\mathrm{x}}{\log \mathrm{x}} \cdot \frac{1}{\mathrm{x}}+\log (\log \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\frac{1+\log (\log x) \cdot(\log x)}{\log x}\right]$
$\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]$
For, $v = x^{\log x}$​​​​​​​
Taking log on both sides, we get
$\log v = \log (x^{\log x})$
$\Rightarrow \log v = \log x. \log x$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[(\log x)^{2}\right]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=2 \cdot \log \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d v}{d x}=v\left[2 \cdot \frac{\log x}{x}\right]$
$\Rightarrow \frac{d v}{d x}=x^{\log x}\left[2 \cdot \frac{\log x}{x}\right]$
$\Rightarrow \frac{d v}{d x}=2 \cdot x^{\log x-1} \cdot \log x$
Because,$ y = u + v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 \cdot x^{\log x-1} \cdot \log x$

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