Find the value of $\frac{d y}{d x}$, if $y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}}$.
We have, $y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}}\quad\text{.... (i)}$
Taking $u=x^{\tan x}$ and $v=\sqrt{\frac{x^2+1}{2}}$
$$\begin{aligned} \log u & =\tan x \log x \quad\text{.... (ii)}\\ \text{and}\quad v^2 & =\frac{x^2+1}{2}\quad\text{.... (iii)} \end{aligned}$$
$$\begin{aligned} &\text { On, differentiating Eq. (ii) w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{1}{u} \cdot \frac{d u}{d x} & =\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^2 x \\ \Rightarrow\quad \frac{d u}{d x} & =u\left[\frac{\tan x}{x}+\log x \cdot \sec ^2 x\right] \\ & =x^{\tan x}\left[\frac{\tan x}{x}+\log x \cdot \sec ^2 x\right]\quad\text{.... (iv)} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Also, differentiating Eq. (iii) w.r.t. } x \text {, we get }\\ &2 v \cdot \frac{d v}{d x}=\frac{1}{2}(2 x) \Rightarrow \frac{d v}{d x}=\frac{1}{4 v} \cdot(2 x) \end{aligned}$$
$\Rightarrow \quad \frac{d v}{d x}=\frac{1}{4 \cdot \sqrt{\frac{x^2+1}{2}}} \cdot 2 x=\frac{x \cdot \sqrt{2}}{2 \sqrt{x^2+1}}$
$\Rightarrow \quad \frac{d v}{d x}=\frac{x}{\sqrt{2\left(x^2+1\right)}}\quad\text{.... (v)}$
$$\begin{aligned} \text { Now, } \quad y & =u+v \\ \therefore \quad \frac{d y}{d x} & =\frac{d u}{d x}+\frac{d v}{d x} \\ & =x^{\tan x}\left[\frac{\tan x}{x}+\log x \cdot \sec ^2 x\right]+\frac{x}{\sqrt{2\left(x^2+1\right)}} \end{aligned}$$
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