To find the derivative of the given function:

$g(x) = \frac{5x^3}{e^x}$

we can use the quotient rule. The quotient rule states that if $g(x) = \frac{f(x)}{h(x)}$, then:

$g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2}$

For this problem:

• $f(x) = 5x^3$ and $h(x) = e^x$.

Step-by-Step Solution

1. Differentiate $f(x) = 5x^3$:

$f'(x) = 15x^2$

2. Differentiate $h(x) = e^x$:

$h'(x) = e^x$

3. Apply the quotient rule:

$g'(x) = \frac{(15x^2)(e^x) - (5x^3)(e^x)}{(e^x)^2}$

4. Simplify the numerator:

$g'(x) = \frac{e^x(15x^2 - 5x^3)}{e^{2x}}$

5. Further simplification:

Since there is an $e^x$ in both the numerator and the denominator, we can cancel one $e^x$ term:

$g'(x) = \frac{15x^2 - 5x^3}{e^x}$

6. Factor out a common term in the numerator:

$g'(x) = \frac{5x^2(3 - x)}{e^x}$

Thus, the derivative of the function is:

$g'(x) = \frac{5x^2(3 - x)}{e^x}$

Would you like further clarification or have any questions?

Here are 5 related questions to consider:

1. How does the quotient rule differ from the product rule in differentiation?
2. What is the significance of the exponential function $e^x$ in calculus?
3. How would the derivative change if the numerator were a different polynomial?
4. Can the expression $\frac{5x^2(3 - x)}{e^x}$ be simplified further?
5. How would you find the critical points of the function $g(x)$ using the derivative?

Tip: Always look for common factors in the numerator when simplifying derivatives; it can make the expression more manageable!