# Find the values

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

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

Solution:

Here,

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

Let

$u=(\log x)^{x}$, and $v=x^{\log x}$

$\therefore \mathrm{y}=\mathrm{u}+\mathrm{v}$

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$           ....(1)

$u=(\log x)^{x}$

$\log u=\log \left[(\log x)^{x}\right]$

$\log u=x \log (\log x)$

Differentiating both sides with respect to $x$, we get

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\log x)+x \frac{d}{d x}[\log (\log x)]$

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

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

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

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

$\frac{d u}{d x}=(\log x)^{x-1}[1+\{\log (\log x) \times \log x\}] \ldots \ldots$ ....(2)

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

$\frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x)$

$\frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x}$

$\frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x}$

$\frac{d v}{d x}=2 x^{\log x-1} \cdot \log x \ldots \ldots$  (3)

Therefore from (i), (ii), (iii), we get

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