Let `f(x)` be a polynomial in `x` . Then the second derivative of `f(e^x)wdotrdottdotxi s` `f^(e^x)dote^x+f^(prime)(e^x)` `f^(e^x)dote^(2x)+f^(prime)(e^x)dote^(2x)` `f^(e^x)e^(2x)` (d) `f^(e^x)dote^(2x)+f^(prime)(e^x)dote^x`

Options

A. `f ''(e^(x))e ^(x)+f (e ^(x))`
B. `f ''(e^(x))e ^(2x)+f '(e ^(x))e ^(2n)`
C. `f '' (e ^(x)) e ^(x)+ f'(e^(x)) e ^(2x)`
D. `f ''(e^(x))e ^(2x) +e ^(x) f'(e^(x))`

Answer

To solve the problem, we need to find the second derivative of the function \( f(e^x) \) with respect to \( x \). Let's go through the steps systematically.

### Step 1: Define the function
Let \( y = f(e^x) \).

### Step 2: First derivative
To find the first derivative \( \frac{dy}{dx} \), we use the chain rule. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.

1. Differentiate \( f(e^x) \):
\[ \frac{dy}{dx} = f'(e^x) \cdot \frac{d}{dx}(e^x) \]
2. The derivative of \( e^x \) is \( e^x \):
\[ \frac{dy}{dx} = f'(e^x) \cdot e^x \]

### Step 3: Second derivative
Now, we need to differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \).

1. Differentiate \( \frac{dy}{dx} = f'(e^x) \cdot e^x \):
Using the product rule, which states that \( (uv)' = u'v + uv' \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(f'(e^x)) \cdot e^x + f'(e^x) \cdot \frac{d}{dx}(e^x) \]

2. Now, we need to differentiate \( f'(e^x) \) using the chain rule again:
\[ \frac{d}{dx}(f'(e^x)) = f''(e^x) \cdot \frac{d}{dx}(e^x) = f''(e^x) \cdot e^x \]

3. Substitute back into the second derivative:
\[ \frac{d^2y}{dx^2} = f''(e^x) \cdot e^x \cdot e^x + f'(e^x) \cdot e^x \]
\[ = e^{2x} f''(e^x) + e^x f'(e^x) \]

### Final Result
Thus, the second derivative of \( f(e^x) \) with respect to \( x \) is:
\[ \frac{d^2y}{dx^2} = e^{2x} f''(e^x) + e^x f'(e^x) \]

### Conclusion
The correct answer corresponds to option (d):
\[ f''(e^x) \cdot e^{2x} + f'(e^x) \cdot e^x \]