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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability4 Marks
Question
Differentiate the function (log x)x + xlog x w.r.t. x.
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Answer

Given: (log x)x + xlog x 
Let y = (log x)x + xlog x 
Let y = u + v
$\Rightarrow$ u = (log x)x and v = xlog 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 = xlog x  
Taking log on both sides, we get
log v = log (xlog 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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