We are asked to find the derivative of the function $f(x) = (-6x + 7)e^{5x - 7}$.

Step-by-step solution:

This function is a product of two parts:

1. $(-6x + 7)$
2. $e^{5x - 7}$

To differentiate, we apply the product rule for derivatives. The product rule states:

$(f \cdot g)' = f' \cdot g + f \cdot g'$

Let:

• $f_1(x) = (-6x + 7)$
• $f_2(x) = e^{5x - 7}$

Now, we calculate the derivatives of both parts:

1. Derivative of $f_1(x) = (-6x + 7)$: $f_1'(x) = -6$

2. Derivative of $f_2(x) = e^{5x - 7}$: Use the chain rule here. The derivative of $e^{u}$ is $e^{u} \cdot u'$, where $u = 5x - 7$. $f_2'(x) = e^{5x - 7} \cdot 5 = 5e^{5x - 7}$

Now, applying the product rule:

$f'(x) = f_1'(x) \cdot f_2(x) + f_1(x) \cdot f_2'(x)$ Substitute the values: $f'(x) = (-6) \cdot e^{5x - 7} + (-6x + 7) \cdot 5e^{5x - 7}$

Simplifying: $f'(x) = -6e^{5x - 7} + 5(-6x + 7)e^{5x - 7}$

Factor out $e^{5x - 7}$: $f'(x) = e^{5x - 7} \left( -6 + 5(-6x + 7) \right)$

Simplify the expression inside the parentheses: $f'(x) = e^{5x - 7} \left( -6 - 30x + 35 \right)$ $f'(x) = e^{5x - 7} \left( -30x + 29 \right)$

Thus, the derivative is: $f'(x) = (-30x + 29) e^{5x - 7}$

Would you like more details, or do you have any questions?

Here are five related questions to deepen your understanding:

1. How does the product rule apply to more complex functions?
2. What is the significance of the chain rule in differentiating exponential functions?
3. Can you differentiate $f(x) = x^2 e^{3x}$ using a similar approach?
4. How does factoring out common terms simplify derivative expressions?
5. How would you interpret the derivative of this function graphically?

Tip: Always look for common terms in your derivative expression to simplify and factorize where possible.