Differentiate the following w.r.t. as indicated :
`x^(2)/(1+x^(2))" w.r.t. "x^(2)`

Answer

To differentiate the function \( \frac{x^2}{1+x^2} \) with respect to \( x^2 \), we can follow these steps:

### Step 1: Define the Variables
Let:
- \( u = \frac{x^2}{1+x^2} \)
- \( v = x^2 \)

### Step 2: Differentiate \( u \) with respect to \( x \)
We will use the quotient rule for differentiation, which states that if \( u = \frac{f(x)}{g(x)} \), then:

\[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]

Here, \( f(x) = x^2 \) and \( g(x) = 1 + x^2 \).

Calculating the derivatives:
- \( f'(x) = 2x \)
- \( g'(x) = 2x \)

Now applying the quotient rule:

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

### Step 3: Simplify \( \frac{du}{dx} \)
Now simplify the numerator:

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

### Step 4: Differentiate \( v \) with respect to \( x \)
Since \( v = x^2 \), we have:

\[ \frac{dv}{dx} = 2x \]

### Step 5: Find \( \frac{du}{dv} \)
Using the chain rule, we know:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

Substituting the derivatives we found:

\[ \frac{du}{dv} = \frac{\frac{2x}{(1+x^2)^2}}{2x} \]

### Step 6: Simplify \( \frac{du}{dv} \)
The \( 2x \) terms cancel out:

\[ \frac{du}{dv} = \frac{1}{(1+x^2)^2} \]

### Final Answer
Thus, the derivative of \( \frac{x^2}{1+x^2} \) with respect to \( x^2 \) is:

\[ \frac{du}{dv} = \frac{1}{(1+x^2)^2} \]