Let `f(x)=(x^(2))/(1-x^(2)),xne0,+-1`, then derivative of f(x) with respect to x, is

Options

A. `(2x)/((1-x^(2))^(2))`
B. `(1)/((2+x^(2))^(3))`
C. `(1)/((1-x^(2))^(2))`
D. `(1)/((2-x^(2))^(2))`

Answer

To find the derivative of the function \( f(x) = \frac{x^2}{1 - x^2} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then the derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step 1: Identify \( u \) and \( v \)
For our function:
- \( u = x^2 \)
- \( v = 1 - x^2 \)

### Step 2: Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
- \( \frac{du}{dx} = 2x \)
- \( \frac{dv}{dx} = -2x \)

### Step 3: Apply the Quotient Rule
Using the quotient rule, we have:

\[ f'(x) = \frac{(1 - x^2)(2x) - (x^2)(-2x)}{(1 - x^2)^2} \]

### Step 4: Simplify the numerator
Now, we simplify the numerator:

\[ = \frac{(1 - x^2)(2x) + 2x^3}{(1 - x^2)^2} \]

Distributing \( 2x \) in the first term:

\[ = \frac{2x - 2x^3 + 2x^3}{(1 - x^2)^2} \]

The \( -2x^3 \) and \( +2x^3 \) cancel each other out:

\[ = \frac{2x}{(1 - x^2)^2} \]

### Final Result
Thus, the derivative of \( f(x) \) is:

\[ f'(x) = \frac{2x}{(1 - x^2)^2} \]