The derivative of `tan^(-1) ((2x)/(1-x^(2)))` with respect to
` cos^(-1) sqrt(1 - x^(2))` is

Options

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

Explanation

To find the derivative of \( \tan^{-1} \left( \frac{2x}{1-x^2} \right) \) with respect to \( \cos^{-1} \sqrt{1 - x^2} \), we can use the chain rule. Let's denote:

- \( U = \tan^{-1} \left( \frac{2x}{1-x^2} \right) \)
- \( V = \cos^{-1} \sqrt{1 - x^2} \)

We want to find \( \frac{dU}{dV} \).

### Step 1: Differentiate \( U \) with respect to \( x \)

Using the derivative of the inverse tangent function, we have:

\[ \frac{dU}{dx} = \frac{1}{1 + \left( \frac{2x}{1-x^2} \right)^2} \cdot \frac{d}{dx} \left( \frac{2x}{1-x^2} \right) \]

Now, we need to differentiate \( \frac{2x}{1-x^2} \). Using the quotient rule:

\[ \frac{d}{dx} \left( \frac{2x}{1-x^2} \right) = \frac{(1-x^2)(2) - (2x)(-2x)}{(1-x^2)^2} = \frac{2(1-x^2) + 4x^2}{(1-x^2)^2} = \frac{2 + 2x^2}{(1-x^2)^2} = \frac{2(1+x^2)}{(1-x^2)^2} \]

Substituting this back, we have:

\[ \frac{dU}{dx} = \frac{1}{1 + \left( \frac{2x}{1-x^2} \right)^2} \cdot \frac{2(1+x^2)}{(1-x^2)^2} \]

Next, we simplify \( 1 + \left( \frac{2x}{1-x^2} \right)^2 \):

\[ 1 + \left( \frac{2x}{1-x^2} \right)^2 = 1 + \frac{4x^2}{(1-x^2)^2} = \frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2} = \frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} = \frac{1 + 2x^2 + x^4}{(1-x^2)^2} \]

Thus, we have:

\[ \frac{dU}{dx} = \frac{2(1+x^2)}{(1-x^2)^2} \cdot \frac{(1-x^2)^2}{1 + 2x^2 + x^4} = \frac{2(1+x^2)}{1 + 2x^2 + x^4} \]

### Step 2: Differentiate \( V \) with respect to \( x \)

Using the derivative of the inverse cosine function, we have:

\[ \frac{dV}{dx} = -\frac{1}{\sqrt{1 - \left( \sqrt{1 - x^2} \right)^2}} \cdot \frac{d}{dx} \left( \sqrt{1 - x^2} \right) \]

Now, we differentiate \( \sqrt{1 - x^2} \):

\[ \frac{d}{dx} \left( \sqrt{1 - x^2} \right) = \frac{-x}{\sqrt{1 - x^2}} \]

Thus, we have:

\[ \frac{dV}{dx} = -\frac{1}{\sqrt{1 - (1 - x^2)}} \cdot \left( -\frac{x}{\sqrt{1 - x^2}} \right) = \frac{x}{\sqrt{x^2}} = \frac{x}{|x|} = 1 \text{ (for } x > 0\text{)} \]

### Step 3: Find \( \frac{dU}{dV} \)

Using the chain rule:

\[ \frac{dU}{dV} = \frac{dU/dx}{dV/dx} = \frac{\frac{2(1+x^2)}{1 + 2x^2 + x^4}}{1} = \frac{2(1+x^2)}{1 + 2x^2 + x^4} \]

### Final Answer:

Thus, the derivative of \( \tan^{-1} \left( \frac{2x}{1-x^2} \right) \) with respect to \( \cos^{-1} \sqrt{1 - x^2} \) is:

\[ \frac{2(1+x^2)}{1 + 2x^2 + x^4} \]