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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:
$\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
✓

Answer

Let, $\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
$=\frac{1}{\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\Big(\frac{1+\cos\text{x}}{\sin\text{x}}\Big)\Bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x})-\sin\text{x}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\Bigg]$
[Using quotient rule]
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\bigg[\frac{(1+\cos\text{x})(\cos\text{x})-\sin\text{x}(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{(1+\cos\text{x})}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\Big)=\text{cosec x}$

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