To find the derivative of $f(x) = 5e^{-x} - 6e^{-5x}$, we will use the chain rule.

1. The derivative of $5e^{-x}$:

   • The derivative of $e^{-x}$ is $-e^{-x}$, so the derivative of $5e^{-x}$ is $5 \times (-e^{-x}) = -5e^{-x}$.

2. The derivative of $-6e^{-5x}$:

   • The derivative of $e^{-5x}$ is $-5e^{-5x}$ (by the chain rule), so the derivative of $-6e^{-5x}$ is $-6 \times (-5e^{-5x}) = 30e^{-5x}$.

Thus, the derivative of $f(x)$ is:

$f'(x) = -5e^{-x} + 30e^{-5x}$

Let me know if you need further clarification or more details! Here are some follow-up questions:

1. How does the chain rule apply when taking derivatives of exponential functions?
2. What happens to the function as $x \to \infty$?
3. Can we find the second derivative of this function?
4. How would you solve for the critical points of $f(x)$?
5. What would the behavior of the function look like graphically?

Tip: When dealing with exponential functions in derivatives, remember the basic rule: $\frac{d}{dx}(e^{kx}) = ke^{kx}$, where $k$ is a constant.