# Find $\large \frac{dy}{dx},\normalsize if\;y=\large x^{tan x}+\sqrt{\large\frac{x^2+1}{2}}$

Toolbox:
• If $y=[f(x)]^{g(x)}$,taking $\log$ on both sides we have $\log y=g(x).\log[f(x)]$.
Step 1:
$y=x^{\large\tan x}+\sqrt{\large\frac{x^2+1}{2}}$
Consider $y=x^{\large \tan x}$
Take $\log$ on both sides,
$\log y=\tan x.\log x$
Differentiating w.r.t $x$ on both sides we get,
$\large\frac{1}{y}\frac{dy}{dx}=$$\tan x.\large\frac{1}{x}$$+\log x.\sec^2 x$
Therefore $\large\frac{dy}{dx}=x^{\large \tan x}\bigg[\large\frac{\tan x}{x}$$+\log x.\sec^2 x\bigg]-----(1) Step 2: Consider y=\sqrt{\large\frac{x^2+1}{2}} Square on both sides y^2=\large\frac{x^2+1}{2} 2y^2=x^2+1 Now differentiate w.r.t x on both sides, 2y\large\frac{dy}{dx}=$$2x$
$\Rightarrow \large\frac{dy}{dx}=\frac{x}{y}$
Substituting for $y$ we get,
Therefore $\large\frac{dy}{dx}=\large\frac{x\sqrt 2}{\sqrt{x^2+1}}$-----(2)
Step 3:
Combining equ(1) and equ(2) we get,