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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives4 Marks
Question
Differentiate the following functions with respect to x:$\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}$
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Answer

We have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}\Big)$
Using quotient rule, we get
$\frac{(\text{x}\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\text{x}\cos\text{x})-(\sin\text{x}-\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x}+\cos\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})\Big\{\cos\text{x}-\Big(\frac{\text{dx}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{dx}}{\text{dx}}\Big)\Big\}-(\sin\text{x}-\text{x}\cos\text{x})\Big(\frac{\text{dx}}{\text{dx}}\sin\text{x}+\sin\text{x}\frac{\text{dx}}{\text{dx}}\Big)+\frac{\text{d}}{\text{dx}}\cos\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})(\cos\text{x}+\text{x}\sin\text{x}-\cos\text{x})-(\sin\text{x}-\text{x}\cos\text{x})(\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})\text{x}\sin\text{x}-(\sin\text{x}-\text{x}\cos\text{x})\text{x}\cos\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{\text{x}^2\sin^2\text{x}+\text{x}\sin\text{x}\cos\text{x}-\text{x}\sin\text{x}\cos\text{x}+\text{x}^2\cos^2\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{\text{x}^2(\sin^2\text{x}+\cos^2\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}\ (\because\sin^2\text{x}+\cos^2\text{x}=1)$
$=\frac{\text{x}^2}{\text{x}\sin\text{x}+\cos\text{x}}$

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