Differentiate w.r.t. 'x' : f(x) = `log((a+b sin x)/(a - b sin x))`

Answer

To differentiate the function \( f(x) = \log\left(\frac{a + b \sin x}{a - b \sin x}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the function
Let \( y = f(x) = \log\left(\frac{a + b \sin x}{a - b \sin x}\right) \).

### Step 2: Apply the logarithmic property
Using the property of logarithms, we can rewrite the function as:
\[ y = \log(a + b \sin x) - \log(a - b \sin x) \]

### Step 3: Differentiate both terms
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}[\log(a + b \sin x)] - \frac{d}{dx}[\log(a - b \sin x)] \]

Using the derivative of \( \log(u) \) which is \( \frac{1}{u} \cdot \frac{du}{dx} \), we get:
\[ \frac{dy}{dx} = \frac{1}{a + b \sin x} \cdot (b \cos x) - \frac{1}{a - b \sin x} \cdot (-b \cos x) \]

### Step 4: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{b \cos x}{a + b \sin x} + \frac{b \cos x}{a - b \sin x} \]

### Step 5: Combine the fractions
To combine these fractions, we find a common denominator:
\[ \frac{dy}{dx} = b \cos x \left(\frac{(a - b \sin x) + (a + b \sin x)}{(a + b \sin x)(a - b \sin x)}\right) \]

This simplifies to:
\[ \frac{dy}{dx} = b \cos x \left(\frac{2a}{a^2 - b^2 \sin^2 x}\right) \]

### Step 6: Final result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{2ab \cos x}{a^2 - b^2 \sin^2 x} \]