#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 31

Answer:  $\frac{d y}{d x}=x^{x}(\log x+1)+x^{\frac{1}{x-2}}(1-\log x)$

Hint: Differentiate the equation taking log on both sides

Given:  $y=x^{x}+x^{\frac{1}{x}}$

Solution:  $y=x^{x}+x^{\frac{1}{x}}$

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

Let $u=x^{x}$

\begin{aligned} &\log u=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=\frac{d}{d x}(x \log x) \end{aligned}

\begin{aligned} &\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d x}{d x} \cdot \log x+x \frac{d}{d x}(\log x) \\\\ &\left.\frac{d u}{d x}=u[\log x+x] \cdot \frac{1}{x}\right] \\\\ &\frac{d u}{d x}=x^{x}[\log x+1] \end{aligned}

Let $v=x^{\frac{1}{x}}$

$\log v=\log x^{\frac{1}{x}}$

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

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

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