Question

If \[ f(x)=\ln\left(\frac{\sin x}{1+\cos x}\right), \] then \(f'(x)\) is equal to:

• A. \( \csc x \)
• B. \( \cot x \)
• C. \( \tan x \)
• D. \( \sec x \)

Hint:

Whenever logarithms involve trigonometric ratios, first try simplifying using identities before differentiating. This often reduces complicated derivatives into standard trigonometric forms.

Answer 1

The Correct Option is A

Concept: To differentiate logarithmic functions involving trigonometric expressions, we use: \[ \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx} \] along with standard trigonometric identities and quotient simplifications.

Step 1: Simplify the logarithmic expression.
Given: \[ f(x)=\ln\left(\frac{\sin x}{1+\cos x}\right) \] Using the identity: \[ \frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x} \] we recognize: \[ \frac{\sin x}{1+\cos x}=\tan\frac{x}{2} \] Thus: \[ f(x)=\ln\left(\tan\frac{x}{2}\right) \]

Step 2: Differentiate the function.
Using: \[ \frac{d}{dx}[\ln u]=\frac{u'}{u} \] we get: \[ f'(x)=\frac{1}{\tan(x/2)}\cdot \sec^2\frac{x}{2}\cdot \frac{1}{2} \] \[ f'(x)=\frac{1}{2}\cdot\frac{\sec^2(x/2)}{\tan(x/2)} \] Now simplify: \[ \frac{\sec^2\theta}{\tan\theta} = \frac{1/\cos^2\theta}{\sin\theta/\cos\theta} = \frac{1}{\sin\theta\cos\theta} \] Hence: \[ f'(x)=\frac{1}{2\sin(x/2)\cos(x/2)} \] Using: \[ 2\sin\frac{x}{2}\cos\frac{x}{2}=\sin x \] we obtain: \[ f'(x)=\frac{1}{\sin x}=\csc x \] Wait, this corresponds to option (A). Let us verify directly using logarithm properties: \[ f(x)=\ln(\sin x)-\ln(1+\cos x) \] Differentiate: \[ f'(x)=\frac{\cos x}{\sin x}-\frac{-\sin x}{1+\cos x} \] \[ f'(x)=\cot x+\frac{\sin x}{1+\cos x} \] Now: \[ \frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x} \] Thus: \[ f'(x)=\frac{\cos x+1-\cos x}{\sin x} \] \[ f'(x)=\frac{1}{\sin x}=\csc x \] Therefore: \[ \boxed{\csc x} \] Hence the correct option is: \[ \boxed{\text{(A)}} \]