The derivative of `tan^(-1)((sqrt(1 + x)-sqrt(1-x))/(sqrt(1 + x)+sqrt(1-x)))` is

Options

A. `sqrt(1 - x^(2))`
B. `(1)/(sqrt(1-x^(2)))`
C. `(1)/(2sqrt(1-x^(2)))`
D. x

Explanation

To find the derivative of the function \( y = \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) \), we can follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent
Let \( z = \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \).

### Step 2: Differentiate Using the Chain Rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{1+z^2} \cdot \frac{dz}{dx} \]
where \( z = \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \).

### Step 3: Differentiate \( z \)
To differentiate \( z \), we will use the quotient rule:
\[ \frac{dz}{dx} = \frac{(\sqrt{1+x} + \sqrt{1-x})\cdot\frac{d}{dx}(\sqrt{1+x} - \sqrt{1-x}) - (\sqrt{1+x} - \sqrt{1-x})\cdot\frac{d}{dx}(\sqrt{1+x} + \sqrt{1-x})}{(\sqrt{1+x} + \sqrt{1-x})^2} \]

### Step 4: Calculate the Derivatives of the Square Roots
The derivatives of the square roots are:
\[ \frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}}, \quad \frac{d}{dx}(\sqrt{1-x}) = -\frac{1}{2\sqrt{1-x}} \]

### Step 5: Substitute the Derivatives into the Quotient Rule
Substituting these derivatives into our expression for \( \frac{dz}{dx} \):
\[ \frac{dz}{dx} = \frac{(\sqrt{1+x} + \sqrt{1-x})\left(\frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}}\right) - (\sqrt{1+x} - \sqrt{1-x})\left(\frac{1}{2\sqrt{1+x}} - \frac{1}{2\sqrt{1-x}}\right)}{(\sqrt{1+x} + \sqrt{1-x})^2} \]

### Step 6: Simplify \( \frac{dz}{dx} \)
This expression can be simplified further, but it may require some algebraic manipulation.

### Step 7: Substitute Back into the Derivative of \( y \)
Once you have \( \frac{dz}{dx} \), substitute it back into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{1+z^2} \cdot \frac{dz}{dx} \]

### Final Expression
Finally, you can express the derivative in terms of \( x \) by substituting back \( z \) and simplifying.

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