To solve this problem, let's go through the steps systematically.

We are given:

1. $x$ and $y$ are functions of time $t$.
2. $x = 4$, $y = 2$, and $\frac{dy}{dt} = -3$.
3. The relationship $xy = x + y$.

We are asked to find $\frac{dx}{dt}$ when the values for $x$, $y$, and $\frac{dy}{dt}$ are as given above.

Step 1: Differentiate the Given Equation

The equation is: $xy = x + y$

Differentiating both sides with respect to $t$ using the product rule on the left side: $\frac{d}{dt}(xy) = \frac{d}{dt}(x + y)$

This yields: $x \frac{dy}{dt} + y \frac{dx}{dt} = \frac{dx}{dt} + \frac{dy}{dt}$

Step 2: Substitute the Known Values

We can substitute $x = 4$, $y = 2$, and $\frac{dy}{dt} = -3$ into the differentiated equation.

So, we have: $4(-3) + 2 \frac{dx}{dt} = \frac{dx}{dt} - 3$

Simplifying this: $-12 + 2 \frac{dx}{dt} = \frac{dx}{dt} - 3$

Step 3: Solve for $\frac{dx}{dt}$

Rearrange the equation to isolate $\frac{dx}{dt}$: $2 \frac{dx}{dt} - \frac{dx}{dt} = -3 + 12$ $\frac{dx}{dt} = 9$

Answer

Thus, the value of $\frac{dx}{dt}$ is: $\frac{dx}{dt} = 9.0$

Would you like further details on any step, or do you have questions?

Here are five related questions to expand your understanding:

1. What is the significance of using implicit differentiation in related rates problems?
2. How would the solution change if $\frac{dy}{dt}$ had a different value?
3. What if the equation were $xy = x - y$ instead; how would that affect the differentiation?
4. How do we apply the product rule in related rates problems involving multiple variables?
5. What are some common applications of related rates in real-world scenarios?

Tip: In related rates problems, carefully apply differentiation rules and always substitute known values only after differentiating.