Differentiate the following functions with respect to x :

Let $y=\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$

$\Rightarrow y=a+b$

where $\mathrm{a}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}} ; \mathrm{b}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where $\mathrm{a}$ and $\mathrm{u}$ are any variables $\}$

$a=\left(x+\frac{1}{x}\right)^{x}$

Taking log both the sides:

$\Rightarrow \log a=\log \left(x+\frac{1}{x}\right)^{x}$

$\Rightarrow \log a=x \log \left(x+\frac{1}{x}\right)$

$\left\{\log x^{a}=a \log x\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log a)}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x} \log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\mathrm{x} \times \frac{\mathrm{d}\left(\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right)}{\mathrm{dx}}+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right) \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=x \times \frac{1}{x+\frac{1}{x}} \frac{d\left(x+\frac{1}{x}\right)}{d x}+\log \left(x+\frac{1}{x}\right)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{x}{\frac{x^{2}+1}{x}}\left\{\frac{d x}{d x}+\frac{d\left(\frac{1}{x}\right)}{d x}\right\}+\log \left(x+\frac{1}{x}\right)$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{x}{\frac{x^{2}+1}{x}}\left\{\frac{d x}{d x}+\frac{d\left(\frac{1}{x}\right)}{d x}\right\}+\log \left(x+\frac{1}{x}\right)$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where a and $\mathrm{u}$ are any variables $\}$

$\Rightarrow \frac{1}{\mathrm{a}} \frac{\mathrm{da}}{\mathrm{dx}}=\frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+1}\left\{1+\left(-\frac{1}{\mathrm{x}^{2}}\right)\right\}+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$

$\left\{\frac{\mathrm{d}\left(\mathrm{u}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{nu}^{\mathrm{n}-1} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\mathrm{a}\left\{\frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+1}\left\{1-\frac{1}{\mathrm{x}^{2}}\right\}+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right\}$

Put the value of $\mathrm{a}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}:$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}\left\{\frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+1}\left\{1-\frac{1}{\mathrm{x}^{2}}\right\}+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right\}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}\left\{\frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+1}-\frac{1}{\mathrm{x}^{2}+1}+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right\}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}\left\{\frac{\mathrm{x}^{2}-1}{\mathrm{x}^{2}+1}+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right\}$

$\mathrm{b}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$

Taking log both the sides:

$\Rightarrow \log a=\log e^{\sin x}$

$\Rightarrow \log a=\sin x \log e$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

$\Rightarrow \log a=\sin x\{\log e=1\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}$

$\Rightarrow \frac{1}{\mathrm{a}} \frac{\mathrm{da}}{\mathrm{dx}}=\cos \mathrm{x}$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}\right\}$

$\Rightarrow \frac{\mathrm{d} \mathrm{a}}{\mathrm{dx}}=\mathrm{a}(\cos \mathrm{x})$

Put the value of $a=e^{\sin x}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=e^{\sin x} \cos x$

$b=(\tan x)^{x}$

Taking log both the sides:

$\Rightarrow \log \mathrm{b}=\log (\tan x)^{x}$

$\Rightarrow \log \mathrm{b}=x \log (\tan x)$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{x} \log (\tan \mathrm{x}))}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\mathrm{x} \times \frac{\mathrm{d}(\log (\tan \mathrm{x}))}{\mathrm{dx}}+\log (\tan \mathrm{x}) \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{x} \times \frac{1}{\tan \mathrm{x}} \frac{\mathrm{d}(\tan \mathrm{x})}{\mathrm{dx}}+\log (\tan \mathrm{x})$

$\left\{\frac{d(\tan x)}{d x}=\sec ^{2} x\right\}$

$\Rightarrow \frac{1}{b} \frac{d b}{d x}=\frac{x}{\tan x}\left(\sec ^{2} x\right)+\log (\tan x)$

$\Rightarrow \frac{1}{b} \frac{d b}{d x}=\frac{x \cos x}{\sin x}\left(\frac{1}{\cos ^{2} x}\right)+\log (\tan x)$

$\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{db}}{\mathrm{dx}}=\frac{\mathrm{x}}{\sin \mathrm{x}}\left(\frac{1}{\cos \mathrm{x}}\right)+\log (\tan \mathrm{x})$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{b}\left\{\frac{\mathrm{x}}{\sin \mathrm{x} \cos \mathrm{x}}+\log (\tan \mathrm{x})\right\}$

Put the value of $b=(\tan x)^{x}$ :

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=(\tan \mathrm{x})^{\mathrm{x}}\left\{\frac{\mathrm{x}}{\sin \mathrm{x} \cos \mathrm{x}}+\log (\tan \mathrm{x})\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\Rightarrow \frac{d y}{d x}=e^{\sin x} \cos x+(\tan x)^{x}\left\{\frac{x}{\sin x \cos x}+\log (\tan x)\right\}$