Differentiate `logsqrt((1+sinx)/(1-sinx))` with respect to `x` :

Answer

To differentiate the function \( y = \log \sqrt{\frac{1 + \sin x}{1 - \sin x}} \) with respect to \( x \), we can follow these steps:

### Step 1: Simplify the expression
We start by simplifying the expression inside the logarithm. The square root can be expressed as an exponent of \( \frac{1}{2} \):
\[ y = \log \left( \frac{1 + \sin x}{1 - \sin x} \right)^{\frac{1}{2}} = \frac{1}{2} \log \left( \frac{1 + \sin x}{1 - \sin x} \right) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \log(a/b) = \log a - \log b \), we can further simplify:
\[ y = \frac{1}{2} \left( \log(1 + \sin x) - \log(1 - \sin x) \right) \]

### Step 3: Differentiate using the chain rule
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{d}{dx} \log(1 + \sin x) - \frac{d}{dx} \log(1 - \sin x) \right) \]
Using the derivative of \( \log u \) which is \( \frac{1}{u} \frac{du}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{1 + \sin x} \cdot \cos x - \frac{1}{1 - \sin x} \cdot (-\cos x) \right) \]

### Step 4: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x} \right) \]
Combining the fractions:
\[ \frac{dy}{dx} = \frac{\cos x}{2} \left( \frac{1 - \sin x + 1 + \sin x}{(1 + \sin x)(1 - \sin x)} \right) = \frac{\cos x}{2} \cdot \frac{2}{\cos^2 x} \]
Since \( (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x \).

### Step 5: Final simplification
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{\cos x} = \sec x \]

### Final Answer
Thus, the derivative of \( y = \log \sqrt{\frac{1 + \sin x}{1 - \sin x}} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \sec x \]