`e^((x^2)//(1+x^2))`

Answer

To find the derivative of the function \( f(x) = e^{\frac{x^2}{1+x^2}} \), we will apply the chain rule and the quotient rule. Here’s a step-by-step solution:

### Step 1: Identify the function and its components
We have:
\[ f(x) = e^{g(x)} \]
where
\[ g(x) = \frac{x^2}{1+x^2} \]

### Step 2: Differentiate using the chain rule
The derivative of \( f(x) \) with respect to \( x \) is given by:
\[ f'(x) = e^{g(x)} \cdot g'(x) \]

### Step 3: Differentiate \( g(x) \) using the quotient rule
To differentiate \( g(x) = \frac{x^2}{1+x^2} \), we use the quotient rule:
\[ g'(x) = \frac{(u'v - uv')}{v^2} \]
where \( u = x^2 \) and \( v = 1 + x^2 \).

Calculating \( u' \) and \( v' \):
- \( u' = 2x \)
- \( v' = 2x \)

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

### Step 4: Simplify \( g'(x) \)
Now we simplify the numerator:
\[ g'(x) = \frac{2x(1+x^2) - 2x^3}{(1+x^2)^2} \]
\[ = \frac{2x + 2x^3 - 2x^3}{(1+x^2)^2} \]
\[ = \frac{2x}{(1+x^2)^2} \]

### Step 5: Substitute \( g'(x) \) back into \( f'(x) \)
Now substituting \( g'(x) \) back into the expression for \( f'(x) \):
\[ f'(x) = e^{g(x)} \cdot g'(x) = e^{\frac{x^2}{1+x^2}} \cdot \frac{2x}{(1+x^2)^2} \]

### Final Result
Thus, the derivative of the function is:
\[ f'(x) = e^{\frac{x^2}{1+x^2}} \cdot \frac{2x}{(1+x^2)^2} \]

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