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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}},$ if $\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}.$
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Answer

We have, $\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}$
Let $\text{u}=\text{x}^{\tan\text{x}}$ and $\text{v}=\sqrt{\frac{\text{x}^2+1}{2}}$
$\therefore\ \log\text{u}\log\text{x}^{\tan\text{x}}=\tan\text{x}\log\text{x}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\tan\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\tan\text{x}}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}\Big]$
$=\text{x}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}\Big]$
Now, $\text{v}=\sqrt{\frac{\text{x}^2+1}{2}}$
Differentiating w.r.t. x, we get
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2}\Big(\frac{\text{x}^2+1}{2}\Big)^{\frac{-1}{2}}\frac{2\text{x}}{2}=\frac{\text{x}}{2}\sqrt{\frac{2}{\text{x}^2+1}}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$
Now, $\text{y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}$ $=\text{x}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}\Big]+\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$

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