https://file.aiquickdraw.com/mathbot/file/1729619383473q1o1vimi.jpg

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.

Assume that x = x(t) and y = y(t). Let y = x^3 + 4 and dx/dt = 2 when x = 5. Find dy/dt when x = 5.

Learn how to apply the chain rule to find dy/dt for the equation y = x^3 + 4, given that dx/dt = 2 when x = 5. This problem involves differentiation and related rates, suitable for students in Grades 10-12.

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