If `y=e^(x^(2)/(1+x^(2)))`, then `(dy)/(dx)=`

Options

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

Answer

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

### Step 1: Identify the outer and inner functions
We have:
\[ y = e^{f(x)} \]
where
\[ f(x) = \frac{x^2}{1+x^2} \]

### Step 2: Differentiate using the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = e^{f(x)} \cdot \frac{df}{dx} \]

### Step 3: Differentiate \( f(x) \) using the quotient rule
To find \( \frac{df}{dx} \), we apply the quotient rule:
If \( u = x^2 \) and \( v = 1 + x^2 \), then:
\[ \frac{df}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 2x \)
- \( \frac{dv}{dx} = 2x \)

Now substituting into the quotient rule:
\[ \frac{df}{dx} = \frac{(1+x^2)(2x) - (x^2)(2x)}{(1+x^2)^2} \]

### Step 4: Simplify the expression
Now simplify the numerator:
\[ (1+x^2)(2x) - (x^2)(2x) = 2x + 2x^3 - 2x^3 = 2x \]

Thus, we have:
\[ \frac{df}{dx} = \frac{2x}{(1+x^2)^2} \]

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

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