Let f be a differentiable function such that f(1) = 2 and f'(x) = f (x) for all `x in R`. If h(x)=f(f(x)), then h'(1) is equal to

Options

A. 4e
B. `4e^2`
C. 2e
D. `2e^2`

Answer

To solve the problem step by step, we will follow the given information and derive the required value of \( h'(1) \).

### Step 1: Understand the function \( f \)
We know that \( f'(x) = f(x) \). This is a differential equation whose general solution is of the form:
\[ f(x) = Ce^x \]
where \( C \) is a constant.

### Step 2: Use the initial condition
We are given that \( f(1) = 2 \). Substituting \( x = 1 \) into the general solution:
\[ f(1) = Ce^1 = Ce = 2 \]
From this, we can solve for \( C \):
\[ C = \frac{2}{e} \]
Thus, the specific form of \( f(x) \) is:
\[ f(x) = \frac{2}{e} e^x = 2e^{x-1} \]

### Step 3: Define the function \( h(x) \)
We have \( h(x) = f(f(x)) \). First, we need to find \( f(f(x)) \):
\[ f(x) = 2e^{x-1} \]
Now substituting \( f(x) \) into itself:
\[ f(f(x)) = f(2e^{x-1}) = 2e^{(2e^{x-1}) - 1} \]

### Step 4: Differentiate \( h(x) \)
To find \( h'(x) \), we will use the chain rule. We have:
\[ h(x) = f(f(x)) \]
Using the chain rule:
\[ h'(x) = f'(f(x)) \cdot f'(x) \]

### Step 5: Find \( f'(x) \)
Since \( f'(x) = f(x) \), we have:
\[ f'(x) = 2e^{x-1} \]

### Step 6: Find \( f'(f(x)) \)
Now we need to evaluate \( f'(f(x)) \):
\[ f'(f(x)) = f(2e^{x-1}) = 2e^{(2e^{x-1}) - 1} \]

### Step 7: Evaluate \( h'(1) \)
Now substituting \( x = 1 \):
1. Calculate \( f(1) \):
\[ f(1) = 2 \]
2. Calculate \( f'(1) \):
\[ f'(1) = f(1) = 2 \]
3. Calculate \( h'(1) \):
\[ h'(1) = f'(f(1)) \cdot f'(1) = f'(2) \cdot 2 \]
4. Calculate \( f'(2) \):
\[ f'(2) = f(2) = 2e^{2-1} = 2e \]
5. Finally, substitute back:
\[ h'(1) = (2e) \cdot 2 = 4e \]

Thus, the final answer is:
\[ \boxed{4e} \]