If ` y = tan ^(-1) ((2x)/(1-x^(2))) , " then " (dy)/(dx)` =____________

Options

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

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), we can follow these steps:

### Step 1: Recognize the function
We start with the function:
\[ y = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

### Step 2: Use the derivative of the inverse tangent
The derivative of \( y = \tan^{-1}(u) \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{2x}{1 - x^2} \).

### Step 3: Differentiate \( u \)
First, we need to find \( \frac{du}{dx} \). We will use the quotient rule for differentiation:
\[ u = \frac{2x}{1 - x^2} \]
Let \( v = 1 - x^2 \). Then,
\[ \frac{du}{dx} = \frac{(1 - x^2)(2) - (2x)(-2x)}{(1 - x^2)^2} \]
This simplifies to:
\[ \frac{du}{dx} = \frac{2(1 - x^2) + 4x^2}{(1 - x^2)^2} = \frac{2 + 2x^2}{(1 - x^2)^2} = \frac{2(1 + x^2)}{(1 - x^2)^2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now we can substitute \( u \) and \( \frac{du}{dx} \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{2x}{1 - x^2} \right)^2} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2} \]

### Step 5: Simplify \( 1 + u^2 \)
Calculating \( 1 + u^2 \):
\[ u^2 = \left( \frac{2x}{1 - x^2} \right)^2 = \frac{4x^2}{(1 - x^2)^2} \]
Thus,
\[ 1 + u^2 = 1 + \frac{4x^2}{(1 - x^2)^2} = \frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2} = \frac{1 - 2x^2 + x^4 + 4x^2}{(1 - x^2)^2} = \frac{1 + 2x^2 + x^4}{(1 - x^2)^2} \]

### Step 6: Final expression for \( \frac{dy}{dx} \)
Now substituting back:
\[ \frac{dy}{dx} = \frac{1}{\frac{1 + 2x^2 + x^4}{(1 - x^2)^2}} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2(1 + x^2)}{1 + 2x^2 + x^4} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2(1 + x^2)}{1 + 2x^2 + x^4} \]