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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability4 Marks
Question
Differentiate the function $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$ w.r.t. x.
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Answer

Given: $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
Let y = $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
Also, let y = u + v
$\Rightarrow \mathrm{u}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^x$ and $\mathrm{v}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$
For, $\mathrm{u}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}$ 
Taking log on both sides, we get
$\log u=\log \left(x+\frac{1}{x}\right)^{x}$ 
Now, differentiate both sides with respect to x
$\frac{d}{d x}(\log u)=\frac{d}{d x}\left[x \cdot \log \left(x+\frac{1}{x}\right)\right]$ 
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right)+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$ 
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right]$ 
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)} \cdot\left(\frac{\mathrm{dx}}{\mathrm{dx}}+\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)\right)+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right]$ 
$\Rightarrow \frac{d u}{d x}=u\left[\frac{x}{\left(x+\frac{1}{x}\right)} \cdot\left(1-\frac{1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right)\right]$
$\Rightarrow \frac{d u}{d x}=u\left[\frac{x}{\left(x+\frac{1}{x}\right)} \cdot\left(\frac{x^{2}-1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right)\right]$ 
$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]$
For, $\mathrm{v}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$ 
Taking log on both sides, we get
$\log v=\log {x}^{\left(1+\frac{1}{x}\right)}$ 
$\Rightarrow \log \mathrm{v}=\left(1+\frac{1}{\mathrm{x}}\right) \cdot \log \mathrm{x}$ 
Now, differentiate both sides with respect to x
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\left(1+\frac{1}{x}\right) \cdot \log x\right]$ 
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(1+\frac{1}{\mathrm{x}}\right)+\left(1+\frac{1}{\mathrm{x}}\right) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$ 
$\Rightarrow \frac{d V}{d x}=v\left[\log x \cdot\left(0-\frac{1}{x^{2}}\right)+\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}\right]$ 
$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[-\frac{\log x}{x^{2}}+\left(\frac{1}{x}+\frac{1}{x^{2}}\right)\right]$ 
$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{-\log x+x+1}{x^{2}}\right]$ 
$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right]$ 
Because, y = u + v
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$ 
$\Rightarrow \frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right]$

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