If `y=log((sin x)/(1+cos x)),"then "(dy)/(dx)`

Options

A. `secx `
B. ` -secx `
C. `cosecx `
D. ` -cosecx `

Answer

To find the derivative of the function \( y = \log\left(\frac{\sin x}{1 + \cos x}\right) \), we will apply the properties of logarithms and differentiation.

### Step-by-Step Solution

1. **Apply the Logarithmic Property**:
\[ y = \log\left(\frac{\sin x}{1 + \cos x}\right) = \log(\sin x) - \log(1 + \cos x) \]

2. **Differentiate Using the Chain Rule**:
We will differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}[\log(\sin x)] - \frac{d}{dx}[\log(1 + \cos x)] \]

3. **Differentiate Each Logarithmic Term**:
- For \( \log(\sin x) \):
\[ \frac{d}{dx}[\log(\sin x)] = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x \]
- For \( \log(1 + \cos x) \):
\[ \frac{d}{dx}[\log(1 + \cos x)] = \frac{1}{1 + \cos x} \cdot (-\sin x) = -\frac{\sin x}{1 + \cos x} \]

4. **Combine the Derivatives**:
Now, substituting back into the derivative:
\[ \frac{dy}{dx} = \cot x + \frac{\sin x}{1 + \cos x} \]

5. **Simplify the Expression**:
We can rewrite \( \cot x \) as \( \frac{\cos x}{\sin x} \):
\[ \frac{dy}{dx} = \frac{\cos x}{\sin x} + \frac{\sin x}{1 + \cos x} \]

6. **Find a Common Denominator**:
The common denominator for the two fractions is \( \sin x(1 + \cos x) \):
\[ \frac{dy}{dx} = \frac{\cos x(1 + \cos x) + \sin^2 x}{\sin x(1 + \cos x)} \]

7. **Use the Pythagorean Identity**:
Recall that \( \sin^2 x + \cos^2 x = 1 \):
\[ \frac{dy}{dx} = \frac{\cos x + \cos^2 x + \sin^2 x}{\sin x(1 + \cos x)} = \frac{\cos x + 1}{\sin x(1 + \cos x)} \]

8. **Final Simplification**:
The numerator simplifies to \( 1 + \cos x \):
\[ \frac{dy}{dx} = \frac{1 + \cos x}{\sin x(1 + \cos x)} = \frac{1}{\sin x} = \csc x \]

### Final Answer
\[ \frac{dy}{dx} = \csc x \]