The problem asks to find the second derivative of $y = e^{-x}$.

Let's work through it step by step:

Step 1: First Derivative

Given: $y = e^{-x}$ Differentiate $y$ with respect to $x$ using the chain rule:

$y' = \frac{d}{dx}(e^{-x}) = e^{-x} \cdot (-1) = -e^{-x}$

Step 2: Second Derivative

Now, differentiate $y' = -e^{-x}$ again with respect to $x$:

$y'' = \frac{d}{dx}(-e^{-x}) = -e^{-x} \cdot (-1) = e^{-x}$

Final Answer:

$y'' = e^{-x}$

Let me know if you'd like any more details or have further questions.

Related Questions:

1. What is the third derivative of $y = e^{-x}$?
2. How does the chain rule apply when differentiating exponential functions?
3. What is the general form of the $n$-th derivative of $e^{-x}$?
4. How does the derivative of $e^{-x}$ compare to that of $e^{x}$?
5. What happens if we differentiate $e^{kx}$, where $k$ is a constant?

Tip:

When differentiating exponentials of the form $e^{kx}$, always apply the chain rule, multiplying by the derivative of the exponent.