If for a continuous function `f,f(0)=f(1)=0,f^(prime)(1)=2a n dy(x)=f(e^x)e^(f(x))` , then `y^(prime)(0)` is equal to a. 1 b. 2 c. 0 d. none of these

Options

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

Explanation

To solve the problem step by step, we start with the given function and apply the necessary differentiation rules.

### Step 1: Write down the function
We are given:
\[ y(x) = f(e^x) \cdot e^{f(x)} \]

### Step 2: Differentiate using the product rule
To find \( y'(x) \), we will use the product rule:
\[ y' = u'v + uv' \]
where \( u = f(e^x) \) and \( v = e^{f(x)} \).

1. Differentiate \( u = f(e^x) \):
- By the chain rule, \( u' = f'(e^x) \cdot e^x \).

2. Differentiate \( v = e^{f(x)} \):
- Again by the chain rule, \( v' = e^{f(x)} \cdot f'(x) \).

Putting it all together:
\[ y'(x) = f'(e^x) \cdot e^x \cdot e^{f(x)} + f(e^x) \cdot e^{f(x)} \cdot f'(x) \]

### Step 3: Evaluate \( y'(0) \)
Now, we need to evaluate \( y'(0) \):
\[ y'(0) = f'(e^0) \cdot e^0 \cdot e^{f(0)} + f(e^0) \cdot e^{f(0)} \cdot f'(0) \]

Since \( e^0 = 1 \) and \( f(0) = 0 \):
\[ y'(0) = f'(1) \cdot 1 \cdot e^0 + f(1) \cdot e^0 \cdot f'(0) \]
\[ y'(0) = f'(1) + f(1) \cdot f'(0) \]

### Step 4: Substitute known values
From the problem statement, we know:
- \( f(1) = 0 \)
- \( f'(1) = 2 \)

Substituting these values:
\[ y'(0) = 2 + 0 \cdot f'(0) \]
\[ y'(0) = 2 \]

### Conclusion
Thus, the value of \( y'(0) \) is:
\[ \boxed{2} \]