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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{x}-\text{e}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}},$ find $\frac{\text{dy}}{\text{dx}}$
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Answer

$\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}$
$\text{y}=\text{e}^{\tan\text{x}\log\text{x}}+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\text{x})+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2}\log\Big(\frac{\text{x}^2+1}{2}\Big)\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{1}{2}\times\frac{2}{\text{x}^2+1}\times(\text{x})\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$

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