If `f(x)=e^(x)g(x),g(0)=2,g'(0)=1, then f'(0) ` is

Options

A. 1
B. 3
C. 2
D. 0

Answer

To find \( f'(0) \) for the function \( f(x) = e^x g(x) \), where \( g(0) = 2 \) and \( g'(0) = 1 \), we will use the product rule of differentiation.

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = e^x \) and \( v = g(x) \).

2. **Apply the Product Rule**:
The product rule states that if \( f(x) = u \cdot v \), then:
\[ f'(x) = u'v + uv' \]
Here, \( u' = \frac{d}{dx}(e^x) = e^x \) and \( v' = g'(x) \).

3. **Differentiate \( f(x) \)**:
Using the product rule:
\[ f'(x) = e^x g(x) + e^x g'(x) \]
We can factor out \( e^x \):
\[ f'(x) = e^x (g(x) + g'(x)) \]

4. **Evaluate at \( x = 0 \)**:
Now we need to find \( f'(0) \):
\[ f'(0) = e^0 (g(0) + g'(0)) \]
Since \( e^0 = 1 \), we have:
\[ f'(0) = 1 \cdot (g(0) + g'(0)) \]

5. **Substitute the Known Values**:
We know from the problem that \( g(0) = 2 \) and \( g'(0) = 1 \):
\[ f'(0) = 1 \cdot (2 + 1) = 1 \cdot 3 = 3 \]

### Final Answer:
Thus, \( f'(0) = 3 \).

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