dy/dx of y= log(sinx/1+cosx)

We have, y = \log [\frac{\sin x}{1 + \cos x}] \Rightarrow \frac{d y}{d x} = \frac{1}{\frac{\sin x}{1 + \cos x}} \times \frac{(1 + \cos x) \times \frac{d}{d x} (\sin x) - \sin x \times \frac{d}{d x} (1 + \cos x)}{{(1 + \cos x)}^{2}} \Rightarrow \frac{d y}{d x} = \frac{1 + \cos x}{\sin x} \times \frac{\cos x (1 + \cos x) + \sin^{2} x}{{(1 + \cos x)}^{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{\sin x} \times \frac{\cos x + \cos^{2} x + \sin^{2} x}{(1 + \cos x)} = \frac{1}{\sin x} \times \frac{(1 + \cos x)}{(1 + \cos x)} \Rightarrow \frac{d y}{d x} = cosec x

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