#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 32 maths

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

Hint: Differentiate the equation taking log on both sides

Given: $y=x^{\log x}+(\log x)^{x}$

Solution:

Let’s assume $y_{1}=x^{\log x} \text { and }$

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

Now $y_{1}=x^{\log x}$

$\log y_{1}=\log x \cdot \log x$

Diff w.r.t x

$\frac{1}{y_{1}} \times \frac{d y_{1}}{d x}=\left[\log x \cdot \frac{1}{x}+\log x \cdot \frac{1}{x}\right]$

$\frac{d y_{1}}{d x}=x^{\log x}\left[\frac{2 \log x}{x}\right]$        .............(1)

\begin{aligned} &y_{2}=(\log x)^{x} \\\\ &\log y_{2}=x \log (\log x) \end{aligned}

Diff both sides w.r.t x

$\frac{1}{y_{2}} \frac{d y_{2}}{d x}=\left[x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \cdot 1+\log (\log x)\right]$

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

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