#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 18 sub question (v)

Answer: $\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)$

Hint: $\text { Diff by }\left(x+\frac{1}{x}\right)^{x}$

Given: $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$

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

$y=u+v$

Diff w.r.t x

$\frac{d y}{d x}=\frac{d(u+v)}{d x}$

$=\frac{d u}{d x}+\frac{d v}{d x}$

Calculate $\frac{d u}{d x}, u=\left(x+\frac{1}{x}\right)^{x}$

Take log on both side

\begin{aligned} &\log u=\log \left(x+\frac{1}{x}\right)^{x} \\\\ &\log u=x \log \left(x+\frac{1}{x}\right) \end{aligned}

$\frac{1}{u} \frac{d u}{d x}=\frac{d\left(x \log \left(1+\frac{1}{x}\right)\right.}{d x}$

Use product rule $(u v)^{\prime}=u^{\prime} v+v^{\prime} u$

$\frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\frac{1}{\left(x+\frac{1}{x}\right)} \frac{d}{d x}\left(x+\frac{1}{x}\right) x$

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

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

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

$\frac{d u}{d x}=\left(x+\frac{1}{x}\right)^{x} \cdot\left(\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right)$

Calculate  $\frac{d v}{d x}$

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

Taking $\log v=\left[1+\frac{1}{x}\right] \log x$

Diff w.r.t. x

Use product rule

$\frac{1}{v} \frac{d v}{d x}=\frac{d\left(1+\frac{1}{x}\right)}{d x} \cdot \log x+\frac{d(\log x)}{d x}\left(1+\frac{1}{x}\right)$

$\frac{1}{v} \frac{d v}{d x}=\frac{d(1)}{d x}+\frac{d\left(\frac{1}{x}\right)}{d x} \cdot \log x+\frac{1}{x}\left(1+\frac{1}{x}\right)$

$\frac{1}{v}\left(\frac{d v}{d x}\right)=\frac{-1}{x^{2}} \cdot \log x+\frac{1}{x}\left(1+\frac{1}{x}\right)$

\begin{aligned} &\frac{1}{v}\left(\frac{d v}{d x}\right)=\left(\frac{-\log x+x+1}{x^{2}}\right) \\\\ &\frac{d v}{d x}=v\left(\frac{-\log x+x+1}{x^{2}}\right) \end{aligned}

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

Now   $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

Put value of  $\frac{d u}{d x} \text { and } \frac{d v}{d x}$

$\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)$