Find the value of $\frac{{dy}}{{dx}}$, if $y = {x^{\tan x}} + \sqrt {\frac{{{x^2} + 1}}{2}}$

We have, $y = {x^{\tan x}} + \sqrt {\frac{{{x^2} + 1}}{2}}$ ………(i)

Taking $u = {x^{\tan x}}$ and $v = \sqrt {\frac{{{x^2} + 1}}{2}}$
$\log u = \tan x\log x$ ……..(ii)

and ${v^2} = \frac{{{x^2} + 1}}{2}$ ……….(iii)

On, differentiating Eq. (ii) w.r.t. $x$ , we get
$\frac{1}{u} \cdot \frac{{du}}{{dx}} = \tan x \cdot \frac{1}{x} + \log x \cdot {\sec ^2}x$

$\Rightarrow$ $\frac{{du}}{{dx}} = u\left[ {\frac{{\tan x}}{x} + \log x \cdot {{\sec }^2}x} \right]$

$= {x^{\tan x}}\left[ {\frac{{\tan x}}{x} + \log x \cdot {{\sec }^2}x} \right]$ ……..(iv)

Also, differentiating Eq. (iii) w.r.t. $x$, we get $2v \cdot \frac{{dv}}{{dx}} = \frac{1}{2}(2x) \Rightarrow \frac{{dv}}{{dx}} = \frac{1}{{4v}} \cdot (2x)$

$\Rightarrow$ $\frac{{dv}}{{dx}} = \frac{1}{{4 \cdot \sqrt {\frac{{{x^2} + 1}}{2}} }} \cdot 2x = \frac{{x \cdot \sqrt 2 }}{{2\sqrt {{x^2} + 1} }}$

$\Rightarrow$ $\frac{{dv}}{{dx}} = \frac{x}{{\sqrt {2\left( {{x^2} + 1} \right)} }}$ …….(v)

Now, $y = u + v$

therefore,$\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$
$= {x^{\tan x}}\left[ {\frac{{\tan x}}{x} + \log x \cdot {{\sec }^2}x} \right] + \frac{x}{{\sqrt {2\left( {{x^2} + 1} \right)} }}$