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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the function given in Exercise:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
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Answer

Let $\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
Putting $\text{u}=\text{x}^{\text{x}\cos\text{x}}\text{ and v}=\frac{\text{x}^2+1}{\text{x}^2-1},\text{we have y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=\text{x}^{\text{x}\cos\text{x}}\ \ \Rightarrow\ \log \text{u}=\log\text{x}^{\text{x}\cos\text{x}}\cos \text{x}=\text{x} \cos\text{x} \log\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\cos\text{x}\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}).\cos\text{x}\log\text{x}+\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\log\text{x}+\text{x}\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1.\cos\text{x}\log\text{x}+\text{x}(-\sin\text{x})\log\text{x}+\text{x}\cos\text{x}\frac{1}{\text{x}}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})\ \dots\text{(ii)}$
Again $\text{v}=\frac{\text{x}^2+1}{\text{x}^2-1}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^2-1)2\text{x}-(\text{x}^2+1)2\text{x}}{(\text{x}^2-1)^2}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{2\text{x}^3-2\text{x}-2\text{x}^3-2\text{x}}{(\text{x}^2-1)^2}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{(\text{x}^2-1)^2}\ \dots\ \text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})+\frac{-4\text{x}}{(\text{x}^2-1)^2}$

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