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.