Differentiate the following w.r.t. x :
`tan^(-1)sqrt((1+sinx)/(1-sinx))`

Answer

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

### Step 1: Rewrite the function
We start with the function:
\[ y = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right) \]

### Step 2: Use the chain rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \).

### Step 3: Differentiate \( u \)
Now we need to differentiate \( u \):
\[ u = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \]
Using the quotient rule, we have:
\[ \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1 + \sin x}{1 - \sin x}}} \cdot \frac{d}{dx}\left(\frac{1 + \sin x}{1 - \sin x}\right) \]

### Step 4: Differentiate the quotient
Using the quotient rule for \( \frac{1 + \sin x}{1 - \sin x} \):
\[ \frac{d}{dx}\left(\frac{1 + \sin x}{1 - \sin x}\right) = \frac{(1 - \sin x)(\cos x) - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2} \]
This simplifies to:
\[ \frac{\cos x(1 - \sin x + 1 + \sin x)}{(1 - \sin x)^2} = \frac{2\cos x}{(1 - \sin x)^2} \]

### Step 5: Substitute back into \( \frac{du}{dx} \)
Now substituting back:
\[ \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1 + \sin x}{1 - \sin x}}} \cdot \frac{2\cos x}{(1 - \sin x)^2} = \frac{\cos x}{\sqrt{\frac{1 + \sin x}{1 - \sin x}}(1 - \sin x)^2} \]

### Step 6: Substitute \( u \) back into the derivative of \( y \)
Now substituting \( u \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)^2} \cdot \frac{\cos x}{\sqrt{\frac{1 + \sin x}{1 - \sin x}}(1 - \sin x)^2} \]

### Step 7: Simplify the expression
Now simplify \( 1 + \left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)^2 \):
\[ 1 + \frac{1 + \sin x}{1 - \sin x} = \frac{(1 - \sin x) + (1 + \sin x)}{1 - \sin x} = \frac{2}{1 - \sin x} \]

Thus, we have:
\[ \frac{dy}{dx} = \frac{1 - \sin x}{2} \cdot \frac{\cos x}{\sqrt{\frac{1 + \sin x}{1 - \sin x}}(1 - \sin x)^2} \]

### Final Result
Combining everything, we get:
\[ \frac{dy}{dx} = \frac{\cos x}{2(1 - \sin x) \sqrt{\frac{1 + \sin x}{1 - \sin x}}} \]