`d/(dx)[e^xlog(1+x^2)]=`

Options

A. `e^x[log(1+x^2)+(2x)/(1+x^2)]`
B. `e^x[log(1+x^2)-(2x)/(1+x^2)]`
C. `e^x[log(1+x^2)+(x)/(1+x^2)]`
D. `e^x[log(1+x^2)-(x)/(1+x^2)]`

Answer

To differentiate the function \( y = e^x \log(1 + x^2) \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

In this case, let:
- \( u = e^x \)
- \( v = \log(1 + x^2) \)

### Step 1: Differentiate \( u \) and \( v \)

First, we need to find the derivatives of \( u \) and \( v \).

1. **Differentiate \( u = e^x \)**:
\[ \frac{du}{dx} = e^x \]

2. **Differentiate \( v = \log(1 + x^2) \)**:
Using the chain rule, we have:
\[ \frac{dv}{dx} = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) = \frac{1}{1 + x^2} \cdot 2x = \frac{2x}{1 + x^2} \]

### Step 2: Apply the Product Rule

Now we apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = e^x \cdot \frac{2x}{1 + x^2} + \log(1 + x^2) \cdot e^x \]

### Step 3: Factor out \( e^x \)

We can factor out \( e^x \) from both terms:
\[ \frac{dy}{dx} = e^x \left( \frac{2x}{1 + x^2} + \log(1 + x^2) \right) \]

### Final Result

Thus, the derivative of \( y = e^x \log(1 + x^2) \) is:
\[ \frac{dy}{dx} = e^x \left( \frac{2x}{1 + x^2} + \log(1 + x^2) \right) \]