Differentiate the following functions with respect to x: 1- 1+

Question

Differentiate the following functions with respect to x:
$\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$

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Answer

Let, $\text{y}=\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
$\Rightarrow\ \text{y}=\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)^\frac{1}{2}$
$\Rightarrow\ \text{y}=\frac{1}{2}\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big) \big[\text{Using }\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\frac{1}{2}\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\Big\}$
$=\frac{1}{2}\times\frac{1}{\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(1-\cos\text{x})-(1-\cos\text{x})\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\bigg[\frac{(1+\cos\text{x})(\sin\text{x})-(1-\cos\text{x})(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\Big[\frac{\sin\text{x}+\sin\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x}\cos\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\Big[\frac{2\sin\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{\sin\text{x}}{(1-\cos\text{x})(1+\cos\text{x})}$
$=\frac{\sin\text{x}}{1-\cos^2\text{x}}$
$=\frac{\sin\text{x}}{\sin^2\text{x}}$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\Big)=\text{cosec x}$