#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 11

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

Hint: Diff by $(\log x)^{\log x}$

Given:$(\log x)^{\log x}$

Solution:  Let $y=(\log x)^{\log x}$

Taking log on both sides

\begin{aligned} &\log y=\log (\log x)^{\log x} \\\\ &\log y=\log x \cdot \log (\log x) \end{aligned}                $\left[\because \log \left(a^{b}\right)=b \log a\right]$

Differentiate w.r.t x,

$\frac{1}{y} \frac{d y}{d x}=d \frac{(\log x \cdot \log (\log x)}{d x}$

Use product rule

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

\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log (\log x)+\frac{1}{\log x}+\frac{d(\log x)}{d x} \log x \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log (\log x)+\frac{d(\log x)}{d x} \end{aligned}

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