We are given:

• $y = x^3 + 4$
• $\frac{dx}{dt} = 2$ when $x = 5$

We are asked to find $\frac{dy}{dt}$ when $x = 5$.

Solution:

To find $\frac{dy}{dt}$, we can use the chain rule, which states: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

First, calculate $\frac{dy}{dx}$: $y = x^3 + 4$ $\frac{dy}{dx} = 3x^2$

Now, substitute $x = 5$ into $\frac{dy}{dx}$: $\frac{dy}{dx} = 3(5)^2 = 3 \times 25 = 75$

Now, using the fact that $\frac{dx}{dt} = 2$, we find $\frac{dy}{dt}$: $\frac{dy}{dt} = 75 \times 2 = 150$

Thus, $\frac{dy}{dt} = 150$ when $x = 5$.

Let me know if you want further details or have questions!

Here are 5 related questions:

1. How does the chain rule apply to implicit differentiation?
2. What are some real-world applications of related rates problems like this one?
3. How does the rate of change of $x$ affect the rate of change of $y$?
4. What is the general formula for differentiating a power of $x$?
5. Can you explain how the product rule would work if $y$ were a product of two functions?

Tip: Always ensure that when applying the chain rule, you correctly differentiate with respect to the intermediate variable and then multiply by its rate of change with respect to time.