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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
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Answer

Let $\text{y}=\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
Also, let $\text{u}=\text{x}^{\text{x}\cos\text{x}}\text{ and v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\therefore\ \text{y}=\text{u}+\text{v}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}\cos\text{x}}$
$\Rightarrow\ \log\text{u}=\log(\text{x}^{\text{x}\cos\text{x}})$
$\Rightarrow\log\text{u}=\text{x}\cos\text{x}\log\text{x}$
Diffrerentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}).\cos\text{x}\log\text{x}+\text{x}.\frac{\text{d}}{\text{dx}}(\cos\text{x}).\log\text{x}+\text{x}\cos\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[1.\cos\text{x}.\log\text{x}+\text{x}.(-\sin\text{x})\log\text{x}+\text{x}\cos\text{x}.\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\cos\text{x})-\text{x}\sin\text{x}\log\text{x}\big]\ .....(\text{ii})$
$\text{v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\Rightarrow\log\text{v}=\log(\text{x}^2+1)-\log(\text{x}^2-1)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{2\text{x}}{\text{x}^2+1}-\frac{2\text{x}}{\text{x}^2-1}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{2\text{x}(\text{x}^2-1)-2\text{x}(\text{x}^2+1)}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+1}{\text{x}^2-1}\times\Big[\frac{-4\text{x}}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{(\text{x}^2-1)^2}\ .....(\text{iii})$
From (1), (2) and (3), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\log\text{x})-\text{x}\sin\text{x}\log\text{x}\big]-\frac{4\text{x}}{(\text{x}^2-1)^2}$

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