If for a continuous function ` f,f(0) = f(1) =0 , f '(1) = 2 and g (x) = f (e^x) e^(f(x)) ,` then g'(0) is equal to

Options

A. 1
B. 2
C. 0
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Define the function**:
\[ g(x) = f(e^x) \cdot e^{f(x)} \]

2. **Differentiate \( g(x) \)** using the product rule:
\[ g'(x) = f'(e^x) \cdot e^x \cdot e^{f(x)} + f(e^x) \cdot e^{f(x)} \cdot f'(x) \]

Here, we have:
- The first term comes from differentiating \( f(e^x) \) (using the chain rule).
- The second term comes from differentiating \( e^{f(x)} \) (also using the chain rule).

3. **Evaluate \( g'(0) \)**:
\[ g'(0) = f'(e^0) \cdot e^0 \cdot e^{f(0)} + f(e^0) \cdot e^{f(0)} \cdot f'(0) \]
Since \( e^0 = 1 \), this simplifies to:
\[ g'(0) = f'(1) \cdot 1 \cdot e^{f(0)} + f(1) \cdot e^{f(0)} \cdot f'(0) \]

4. **Substitute known values**:
- From the problem, we know \( f(0) = 0 \), \( f(1) = 0 \), and \( f'(1) = 2 \).
- Therefore, \( e^{f(0)} = e^0 = 1 \).

Substituting these values into the equation:
\[ g'(0) = f'(1) \cdot 1 + 0 \cdot 1 \cdot f'(0) \]
This simplifies to:
\[ g'(0) = 2 + 0 = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]