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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}$
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Answer

Let $\text{y}=\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{2}^\text{x}\cos\text{x}}{(\text{x}^2+3)^3}\Big]$
$=\bigg[\frac{(\text{x}^2+3)^2\frac{\text{d}}{\text{dx}}(2^\text{x}\cos\text{x})-(2^\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^2+3)^2}{\big[(\text{x}^2+3)\big]^2}\bigg]$
[Using quotient rule]
$=\Bigg[\frac{(\text{x}^2+3)^2\Big\{2^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}2^\text{x}\Big\}-(2^\text{x}+3)2(\text{x}^2+3)\frac{\text{d}}{\text{dx}}(\text{x}^2+3)}{(\text{x}^2+3)^4}\Bigg]$
[Using Product rule and chain rule]
$=\bigg[\frac{(\text{x}^2+3)^2\big\{-2^\text{x}\sin\text{x}+\cos\text{x}2^\text{x}\log_\text{e}2\big\}-2(2^\text{x}\cos\text{x})(\text{x}^2+3)(2\text{x})}{(\text{x}^2+3)^4}\bigg]$
$=\bigg[\frac{2^\text{x}(\text{x}^2+3)\big\{(\text{x}^2+3)(\cos\text{x}\log_\text{e}2-\sin\text{x})-4\text{x}\cos\text{x}\big\}}{(\text{x}^2+3)^4}\bigg]$
$=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{2^\text{x}\cos\text{x}}{(\text{x}^2+3)^2}\Big]=\frac{2^\text{x}}{(\text{x}^2+3)^2}\bigg[\cos\text{x}\log_\text{e}2-\sin\text{x}-\frac{4\text{x}\cos\text{x}}{(\text{x}^2+3)}\bigg]$

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