Question:
If $y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$, find $\frac{d y}{d x}$
Solution:
Given $y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$
$y=e^{\tan x \log x}+e^{\frac{1}{2} \log \frac{x^{2}+1}{2}}$
$\frac{d y}{d x}=e^{\tan x \log x} \frac{d}{d x}(\tan x \log x)+e^{\frac{1}{2} \log \frac{x^{2}+1}{2}} \frac{d}{d x}\left(\frac{1}{2} \log \frac{x^{2}+1}{2}\right)$
$\frac{d y}{d x}=x^{\tan x}\left[\frac{\tan x}{x}+\sec ^{2} \log x\right]+\sqrt{\frac{x^{2}+1}{2}}\left(\frac{1}{2} \times \frac{2}{x^{2}+1} \times x\right)$
$\frac{d y}{d x}=x^{\tan x}\left[\frac{\tan x}{x}+\sec ^{2} \log x\right]+\sqrt{\frac{x^{2}+1}{2}}\left(\frac{x}{x^{2}+1}\right)$
$\frac{d y}{d x}=x^{\tan x}\left[\frac{\tan x}{x}+\sec ^{2} \log x\right]+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}$