Question
Differentiate the function $x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1} w.r.t. x.$
✓
Answer
Given: $\mathrm{x}^{\mathrm{x} \cos \mathrm{x}}+\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}-1}$
Let $y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
Let $y = u + v$
$\Rightarrow u = x^{xcos x}$ and $v = \frac{x^{2}+1}{x^{2}-1}$
For, $\mathrm{u}=\mathrm{x}^{\mathrm{x} \cos x}$
Taking $\log$ on both sides, we get
$\log u=\log {x} ^{x \cos x}$
$\Rightarrow \log u=x \cdot \cos x \cdot \log x$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \cos x \cdot \log x]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\cos \mathrm{x} \cdot \log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\mathrm{x} \cdot \log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})+\mathrm{x} \cdot \cos \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}$
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=\mathrm{u}\left[\cos \mathrm{x} . \log \mathrm{x}+\mathrm{x} \cdot \log \mathrm{x}(-\sin \mathrm{x})+\mathrm{x} \cdot \cos \mathrm{x} \cdot\left(\frac{1}{\mathrm{x}}\right)\right]$
$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x \cdot \log x-x \cdot \log x \cdot \sin x+\cos x]$
$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]$
For $, v = \frac{x^{2}+1}{x^{2}-1}$
Takinglog on both sides, we get
$\log \mathrm{v}=\log \left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}-1}\right)$
$\Rightarrow \log v=\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)\right]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\frac{1}{\mathrm{x}^{2}+1} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)-\frac{1}{\mathrm{x}^{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)$
$\Rightarrow \frac{d v}{d x}=v \cdot\left[\frac{1}{x^{2}+1} \cdot(2 x)-\frac{1}{x^{2}-1} \cdot(2 x)\right]$
$\Rightarrow \frac{d v}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{2 x\left(x^{2}-1\right)-2 x\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]$
$\Rightarrow \frac{d v}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{-4 x}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]$
$\Rightarrow \frac{d v}{d x}=\left[\frac{-4 x}{\left(x^{2}-1\right)^{2}}\right]$
Because $, y = u + v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]-\left[\frac{4 x}{\left(x^{2}-1\right)^{2}}\right]$
Let $y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
Let $y = u + v$
$\Rightarrow u = x^{xcos x}$ and $v = \frac{x^{2}+1}{x^{2}-1}$
For, $\mathrm{u}=\mathrm{x}^{\mathrm{x} \cos x}$
Taking $\log$ on both sides, we get
$\log u=\log {x} ^{x \cos x}$
$\Rightarrow \log u=x \cdot \cos x \cdot \log x$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \cos x \cdot \log x]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\cos \mathrm{x} \cdot \log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\mathrm{x} \cdot \log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})+\mathrm{x} \cdot \cos \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}$
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=\mathrm{u}\left[\cos \mathrm{x} . \log \mathrm{x}+\mathrm{x} \cdot \log \mathrm{x}(-\sin \mathrm{x})+\mathrm{x} \cdot \cos \mathrm{x} \cdot\left(\frac{1}{\mathrm{x}}\right)\right]$
$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x \cdot \log x-x \cdot \log x \cdot \sin x+\cos x]$
$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]$
For $, v = \frac{x^{2}+1}{x^{2}-1}$
Takinglog on both sides, we get
$\log \mathrm{v}=\log \left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}-1}\right)$
$\Rightarrow \log v=\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)\right]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\frac{1}{\mathrm{x}^{2}+1} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)-\frac{1}{\mathrm{x}^{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)$
$\Rightarrow \frac{d v}{d x}=v \cdot\left[\frac{1}{x^{2}+1} \cdot(2 x)-\frac{1}{x^{2}-1} \cdot(2 x)\right]$
$\Rightarrow \frac{d v}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{2 x\left(x^{2}-1\right)-2 x\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]$
$\Rightarrow \frac{d v}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{-4 x}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]$
$\Rightarrow \frac{d v}{d x}=\left[\frac{-4 x}{\left(x^{2}-1\right)^{2}}\right]$
Because $, y = u + v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]-\left[\frac{4 x}{\left(x^{2}-1\right)^{2}}\right]$
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