Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives4 Marks
Question
Differentiate the following functions with respect to x:$\frac{\text{x}\sin\text{x}}{1+\cos\text{x}}$
✓
Answer
We have, $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\sin\text{x}}{1+\cos\text{x}}\Big)$
Using Quotient rule, we get
$=\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x})-(\text{x}\sin\text{x})\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}$
$=\frac{(1+\cos\text{x})\Big(\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}+\sin\text{x}\frac{\text{d}}{\text{dx}}\Big)-\text{x}\sin\text{x}(-\sin\text{x})}{(1+\cos\text{x})^2}$ [Used product rule]
$=\frac{(1+\cos\text{x})(\text{x}\cos\text{x}+\sin\text{x})+\text{x}\sin^2\text{x}}{(1+\cos\text{x})^2}$
$=\frac{\text{x}\cos\text{x}+\text{x}\cos^2\text{x}+\sin\text{x}+\sin\text{x}\cos\text{x}+\text{x}\sin^2\text{x}}{(1+\cos\text{x})^2}$
$=\frac{(\text{x}\cos\text{x}+\sin\text{x}+\sin\text{x}\cos\text{x})+\text{x}(\sin^2\text{x}+\cos^2\text{x})}{(1+\cos\text{x})^2}$
$=\frac{\sin\text{x}+\sin\text{x}\cos\text{x}+\text{x}(\cos\text{x}+\sin^2\text{x}+\cos^2\text{x})}{(1+\cos\text{x})^2}$
$=\frac{\sin\text{x}(1+\cos\text{x})+\text{x}(\cos\text{x}+1)}{(1+\cos\text{x})^2}$
$=\frac{(\text{x}+\sin\text{x})(\cos\text{x}+1)}{(1+\cos\text{x})^2}$
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