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Find \frac{d y}{d x}, \text { if } y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}

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Solution

We have, 

y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}
Putting x^{ \tan x}=u\ and\ \sqrt{\frac{x^{2}+1}{2}}=v
u=x ^{\tan x}
Taking log on both sides, we get
\log u=\tan x \log x
Differentiating w.r.t x, we get
\Rightarrow \frac{1}{u} \frac{d u}{d x}=\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x

\\\Rightarrow \frac{d u}{d x}=u\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x\right) \\ \Rightarrow \frac{d u}{d x}=x^{\tan x}\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x\right)

Now, \mathrm{v}=\sqrt{\frac{\mathrm{x}^{2}+1}{2}}
\Rightarrow v=\left(\frac{x^{2}+1}{2}\right)^{1 / 2}
Differentiating w.r.t x, we get
\\ \Rightarrow \frac{d v}{d x}=\frac{1}{2}\left(\frac{x^{2}+1}{2}\right)^{-1 / 2} \cdot \frac{2 x}{2} \\\Rightarrow \frac{d v}{d x}=\frac{x}{2}\left(\frac{2}{x^{2}+1}\right)^{1 / 2}=\frac{x}{2} \sqrt{\frac{2}{x^{2}+1}}
Now,  y=u+v 
\\\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\\Rightarrow \frac{d y}{d x}=x^{\tan x}\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x\right)+\frac{x}{2} \sqrt{\frac{2}{x^{2}+1}}

\\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\operatorname{tanx}}\left(\frac{\tan \mathrm{x}}{\mathrm{x}}+\log \mathrm{x} \cdot \sec ^{2} \mathrm{x}\right)+\frac{\mathrm{x}}{\sqrt{2\left(\mathrm{x}^{2}+1\right)}}

 

Posted by

infoexpert22

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