Let's analyze the function $f(x) = 2 e^{-x^2}$ and find its second derivative, $f''(x)$.

Part 1: Finding $f''(x)$

1. First Derivative $f'(x)$: $f'(x) = \frac{d}{dx}(2 e^{-x^2}) = 2 \cdot (-2x) e^{-x^2} = -4x e^{-x^2}$

2. Second Derivative $f''(x)$: To find $f''(x)$, we differentiate $f'(x)$: $f''(x) = \frac{d}{dx}(-4x e^{-x^2}) = -4 \left( e^{-x^2} + x \frac{d}{dx}(e^{-x^2}) \right)$ Using the chain rule for $e^{-x^2}$: $\frac{d}{dx}(e^{-x^2}) = -2x e^{-x^2}$ So: $f''(x) = -4 \left( e^{-x^2} + x(-2x e^{-x^2}) \right) = -4 \left( e^{-x^2} - 2x^2 e^{-x^2} \right)$ $= -4 e^{-x^2}(1 - 2x^2) = 8x^2 e^{-x^2} - 4 e^{-x^2}$

Thus, the second derivative is: $f''(x) = 8x^2 e^{-x^2} - 4 e^{-x^2}$

Part 2: Finding $f''(0)$

Now, we substitute $x = 0$: $f''(0) = 8(0)^2 e^{0} - 4 e^{0} = 0 - 4 = -4$

Part 3: Finding $f''(1)$

Now, substitute $x = 1$: $f''(1) = 8(1)^2 e^{-1^2} - 4 e^{-1} = 8 e^{-1} - 4 e^{-1} = 4 e^{-1}$

To express this numerically: $f''(1) = 4 \cdot \frac{1}{e} \approx 4 \cdot 0.367879 \approx 1.4716$

Summary of Results

• Part 1: $f''(x) = 8x^2 e^{-x^2} - 4 e^{-x^2}$
• Part 2: $f''(0) = -4$ (Correct)
• Part 3: $f''(1) = 4 e^{-1} \approx 1.4716$ (Correct)

If you need further details or have questions, let me know! Here are some related questions:

1. What is the significance of the second derivative in the context of the original function?
2. How would the behavior of $f''(x)$ change if the coefficient of $e^{-x^2}$ was increased?
3. Can you explain the use of the chain rule in finding derivatives of composite functions?
4. What is the value of $e^{-1}$ and how is it derived?
5. How would you approach finding higher-order derivatives of $f(x)$?

Tip: When finding derivatives of exponential functions, always remember to apply the chain rule carefully!