Find the differential coefficient of the following function with respect to 'x'
`e^((x^2)//(1+x^2))`

Answer

To find the differential coefficient of the function \( y = e^{\frac{x^2}{1+x^2}} \) with respect to \( x \), we will use the chain rule and the quotient rule for differentiation. Here’s a step-by-step solution:

### Step 1: Define the function
Let
\[ y = e^{\frac{x^2}{1+x^2}}. \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we will apply the chain rule. The derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \), where \( u = \frac{x^2}{1+x^2} \).

Thus, we have:
\[ \frac{dy}{dx} = e^{\frac{x^2}{1+x^2}} \cdot \frac{d}{dx}\left(\frac{x^2}{1+x^2}\right). \]

### Step 3: Differentiate \( u = \frac{x^2}{1+x^2} \) using the quotient rule
Let \( u = \frac{x^2}{1+x^2} \). We will use the quotient rule, which states that if \( u = \frac{f(x)}{g(x)} \), then:
\[ \frac{du}{dx} = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}. \]
Here, \( f(x) = x^2 \) and \( g(x) = 1 + x^2 \).

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

Now applying the quotient rule:
\[ \frac{du}{dx} = \frac{(1+x^2)(2x) - (x^2)(2x)}{(1+x^2)^2}. \]

### Step 4: 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 5: Substitute back into the derivative
Now substituting \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = e^{\frac{x^2}{1+x^2}} \cdot \frac{2x}{(1+x^2)^2}. \]

### Final Result
Thus, the differential coefficient of the function is:
\[ \frac{dy}{dx} = e^{\frac{x^2}{1+x^2}} \cdot \frac{2x}{(1+x^2)^2}. \]