https://file.aiquickdraw.com/mathbot/file/17291345652456p69ecb5.jpg

The problem in the image is as follows:

Given:
\[
xy = 5 \quad \text{and} \quad \frac{dy}{dt} = 4
\]
Find \(\frac{dx}{dt}\) when \(x = 5\).

### Solution:

1. Start by differentiating the equation \(xy = 5\) with respect to time \(t\), using implicit differentiation. Recall that both \(x\) and \(y\) are functions of \(t\), so we apply the product rule:

\[
\frac{d}{dt}(xy) = \frac{d}{dt}(5)
\]
Using the product rule for the left-hand side:
\[
\frac{dx}{dt}y + x\frac{dy}{dt} = 0
\]
(since the derivative of a constant, 5, is 0).

2. Now, substitute the given information:
   - \(\frac{dy}{dt} = 4\)
   - When \(x = 5\), solve for \(y\) from the original equation \(xy = 5\): 
   \[
   5y = 5 \quad \Rightarrow \quad y = 1
   \]

So the equation becomes:
\[
\frac{dx}{dt}(1) + 5(4) = 0
\]
\[
\frac{dx}{dt} + 20 = 0
\]
\[
\frac{dx}{dt} = -20
\]

### Final Answer:
\[
\frac{dx}{dt} = -20
\]

Would you like further details on any step? Feel free to ask!

Here are 5 related questions to expand on this concept:
1. What is implicit differentiation and when is it useful?
2. Can you explain the product rule and its applications in calculus?
3. How do you find \(\frac{dx}{dt}\) when both variables are changing with respect to time?
4. If \(xy = 7\) and \(\frac{dy}{dt} = 3\), how would you find \(\frac{dx}{dt}\) when \(x = 7\)?
5. How can you interpret the meaning of \(\frac{dx}{dt}\) in a physical scenario?

**Tip:** Always pay attention to when variables are functions of time, as you’ll often need to apply implicit differentiation in these cases.

Let xy = 5 and dy/dt = 4. Find dx/dt when x = 5.

Learn how to find dx/dt when given xy = 5 and dy/dt = 4 using implicit differentiation. This problem demonstrates the use of the product rule and solving for rates of change. Suitable for students in Grades 10-12.

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