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

Step 1 : Understanding the Question:
The goal is to find the derivative of a composite function consisting of a natural logarithm containing a trigonometric rational expression. Direct differentiation can be complex, so simplifying the trigonometric function first is an elegant strategy.
Step 2 : Key Formulas and Approach:
We use trigonometric half-angle and double-angle formulas to simplify the argument:
\[ \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}, \quad 1 + \cos x = 2\cos^2\frac{x}{2} \] We also apply the standard derivative rule for natural logarithms:
\[ \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \] Using these tools, we will simplify the expression before performing differentiation.
Step 3 : Detailed Solution:

Simplify the internal fraction using trigonometric identities:
\[ \frac{\sin x}{1 + \cos x} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \tan\frac{x}{2} \]
Rewrite the original function in a simpler form:
\[ f(x) = \ln\left(\tan\frac{x}{2}\right) \]
Differentiate the simplified function using the chain rule:
\[ f'(x) = \frac{1}{\tan(x/2)} \cdot \frac{d}{dx}\left(\tan\frac{x}{2}\right) \]
Apply the derivative of the tangent function:
\[ f'(x) = \frac{1}{\tan(x/2)} \cdot \sec^2\frac{x}{2} \cdot \frac{1}{2} \]
Convert the trigonometric terms back to sine and cosine to simplify:
\[ f'(x) = \frac{\cos(x/2)}{\sin(x/2)} \cdot \frac{1}{\cos^2(x/2)} \cdot \frac{1}{2} = \frac{1}{2\sin(x/2)\cos(x/2)} \]
Apply the double-angle formula \( 2\sin(x/2)\cos(x/2) = \sin x \):
\[ f'(x) = \frac{1}{\sin x} = \csc x \]
Step 4 : Final Answer:
The derivative of the function is \( \csc x \), which corresponds to option (A).
\[ \boxed{\text{(A)}} \]