Question

Differentiate \[\log \left( \frac{\sin x}{1 + \cos x} \right)\] ?

Answer 1

\[\text{Let y} = \log\left( \frac{\sin x}{1 + \cos x} \right)\]

\[\text{Differentiate it with respect to x, we get}\]

\[\frac{d y}{d x} = \frac{d}{dx}\log\left( \frac{\sin x}{1 + \cos x} \right)\]

\[ = \frac{1}{\left( \frac{\sin x}{1 + \cos x} \right)} \times \frac{d}{dx}\left( \frac{\sin x}{1 + \cos x} \right) ........\left[ \text{Using chain rule } \right]\]

\[ = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\left( 1 + \cos x \right)\frac{d}{dx}\left( \sin x \right) - \sin x\frac{d}{dx}\left( 1 + \cos x \right)}{\left( 1 + \cos x \right)^2} \right] ........\left[ \text{Using quotient rule} \right]\]

\[ = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\left( 1 + \cos x \right)\left( \cos x \right) - \sin x\left( - \sin x \right)}{\left( 1 + \cos x \right)^2} \right]\]

\[ = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\cos x + \cos^2 x + \sin^2 x}{\left( 1 + \cos x \right)^2} \right]\]

\[ = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\left( 1 + \cos x \right)}{\left( 1 + \cos x \right)^2} \right]\]

\[ = \frac{1}{\sin x}\]

`= "cosec "x`

So, `d/(dx){log((sin x)/(1+cos x))}= "cosec "x`