To solve the problem, let's take the derivative of each term in the expression separately. The expression is:

$\frac{d}{dx} \left( 9e^{-x} - 7e^{4x} - 8e^x \right)$

We'll apply the derivative rules for exponential functions:

1. First Term: $9e^{-x}$

   • Derivative: $\frac{d}{dx} (9e^{-x}) = 9 \cdot \frac{d}{dx}(e^{-x}) = 9 \cdot (-e^{-x}) = -9e^{-x}$

2. Second Term: $-7e^{4x}$

   • Derivative: $\frac{d}{dx} (-7e^{4x}) = -7 \cdot \frac{d}{dx}(e^{4x}) = -7 \cdot (4e^{4x}) = -28e^{4x}$

3. Third Term: $-8e^x$

   • Derivative: $\frac{d}{dx} (-8e^x) = -8 \cdot \frac{d}{dx}(e^x) = -8e^x$

Now, combining these results:

$\frac{d}{dx} \left( 9e^{-x} - 7e^{4x} - 8e^x \right) = -9e^{-x} - 28e^{4x} - 8e^x$

Thus, the answer is:

$-9e^{-x} - 28e^{4x} - 8e^x$

Would you like further clarification on any part of this solution?

Here are five related questions for further exploration:

1. How do we handle derivatives of more complex exponential functions?
2. What are the derivative rules for logarithmic functions?
3. How does the chain rule work with composite exponential functions?
4. What are the applications of derivatives involving exponential functions in real-world scenarios?
5. How do we find the second derivative of the given expression?

Tip: Remember that exponential functions $e^{kx}$ have derivatives $k \cdot e^{kx}$ due to the chain rule.