To find $\frac{dx}{dt}$ given the equation $y^2 + xy - 3x = -1,$ and knowing $\frac{dy}{dt} = 2$ when $x = 1$ and $y = 1$, we’ll use implicit differentiation with respect to $t$.

Step-by-Step Solution

1. Differentiate both sides of the equation with respect to $t$:

   $\frac{d}{dt}(y^2) + \frac{d}{dt}(xy) - \frac{d}{dt}(3x) = \frac{d}{dt}(-1).$

2. Apply the chain rule:

   • For $y^2$, we have: $\frac{d}{dt}(y^2) = 2y \cdot \frac{dy}{dt}.$

   • For $xy$, we use the product rule: $\frac{d}{dt}(xy) = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}.$

   • For $3x$, we get: $\frac{d}{dt}(3x) = 3 \cdot \frac{dx}{dt}.$

   Putting it all together, the differentiated equation is: $2y \cdot \frac{dy}{dt} + \left( \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt} \right) - 3 \cdot \frac{dx}{dt} = 0.$

3. Substitute the known values $x = 1$, $y = 1$, and $\frac{dy}{dt} = 2$:

   $2(1)(2) + \left( \frac{dx}{dt} \cdot 1 + 1 \cdot 2 \right) - 3 \cdot \frac{dx}{dt} = 0.$

   Simplifying each term: $4 + \left( \frac{dx}{dt} + 2 \right) - 3 \cdot \frac{dx}{dt} = 0.$

4. Combine like terms:

   $4 + 2 - 2 \cdot \frac{dx}{dt} = 0.$

   This simplifies further to: $6 - 2 \cdot \frac{dx}{dt} = 0.$

5. Solve for $\frac{dx}{dt}$:

   $2 \cdot \frac{dx}{dt} = 6 \Rightarrow \frac{dx}{dt} = 3.$

Final Answer

$\frac{dx}{dt} = 3.$

Would you like further details or have any questions?

Here are some follow-up questions for deeper understanding:

1. Why did we use implicit differentiation for this problem?
2. How does the product rule apply when differentiating $xy$ with respect to $t$?
3. What are other applications of implicit differentiation in calculus?
4. How would the solution change if $\frac{dy}{dt}$ had a different value?
5. What would happen if we needed to find $\frac{d^2x}{dt^2}$ as well?

Tip: When differentiating implicitly, always remember to apply the chain rule carefully for each term.