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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
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Answer

Consider $\text{y}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2+2)^3\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\sin\text{x})-\text{e}^\text{x}\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)^3}{\big[(\text{x}^2+2)^3\big]^2}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-\text{e}^\text{x}\sin\text{x}3(\text{x}^2+2)^2(2\text{x})}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-6\text{xe}^\text{x}\sin\text{x}(\text{x}^2+2)^2}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^2\big[(\text{x}^2+2)(\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x})-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^6}$
$=\frac{\text{x}^2\text{e}^\text{x}\cos\text{x}+\text{x}^2\sin\text{xe}^\text{x}+2\text{e}^\text{x}\cos\text{x}+2\sin\text{xe}^\text{x}-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
$=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$

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