Let’s solve the two problems from the uploaded image step by step.

Problem 1:

Find the slope of the tangent line to the curve $x^3 + xy + y^3 = 5$ at the point $(-1, 2)$.

Step 1: Differentiate the given equation implicitly

The given equation is: $x^3 + xy + y^3 = 5$ Taking the derivative of both sides with respect to $x$, we get: $\frac{d}{dx}(x^3) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5)$ Using the chain rule and the product rule: $3x^2 + \left( x\frac{dy}{dx} + y \right) + 3y^2\frac{dy}{dx} = 0$

Step 2: Simplify to isolate $\frac{dy}{dx}$

Reorganizing terms: $x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -3x^2 - y$ Factor $\frac{dy}{dx}$ on the left-hand side: $\frac{dy}{dx} (x + 3y^2) = -3x^2 - y$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-3x^2 - y}{x + 3y^2}$

Step 3: Evaluate $\frac{dy}{dx}$ at $(-1, 2)$

Substitute $x = -1$ and $y = 2$: $\frac{dy}{dx} = \frac{-3(-1)^2 - 2}{-1 + 3(2)^2}$ Simplify: $\frac{dy}{dx} = \frac{-3(1) - 2}{-1 + 3(4)}$ $\frac{dy}{dx} = \frac{-3 - 2}{-1 + 12}$ $\frac{dy}{dx} = \frac{-5}{11}$

Thus, the slope of the tangent line is: $\boxed{-\frac{5}{11}}$

Problem 2:

Find $\frac{dy}{dx}$ if $x^2y + y^2x = -2$.

Step 1: Differentiate the given equation implicitly

The given equation is: $x^2y + y^2x = -2$ Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2y) + \frac{d}{dx}(y^2x) = \frac{d}{dx}(-2)$ Using the product rule: $(2x)y + x^2\frac{dy}{dx} + (2y)\frac{dy}{dx}x + y^2 = 0$

Step 2: Simplify to isolate $\frac{dy}{dx}$

Reorganize terms: $2xy + x^2\frac{dy}{dx} + 2xy\frac{dy}{dx} + y^2 = 0$ Group terms involving $\frac{dy}{dx}$: $x^2\frac{dy}{dx} + 2xy\frac{dy}{dx} = -2xy - y^2$ Factor $\frac{dy}{dx}$: $\frac{dy}{dx}(x^2 + 2xy) = -2xy - y^2$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}$

Summary of Results

1. Slope of the tangent line for Problem 1: $\boxed{-\frac{5}{11}}$
2. Expression for $\frac{dy}{dx}$ in Problem 2: $\boxed{\frac{-2xy - y^2}{x^2 + 2xy}}$

Let me know if you’d like further explanations or clarifications!

Related Questions

1. How does implicit differentiation work when both $x$ and $y$ are involved in products?
2. Can the slope of a tangent line be interpreted geometrically for these problems?
3. How would the slope change if the point in Problem 1 were different?
4. Could Problem 2 involve any critical points for $y$ (e.g., $\frac{dy}{dx} = 0$)?
5. How does the product rule simplify differentiation in these scenarios?

Tip: Always isolate $\frac{dy}{dx}$ systematically by factoring and reorganizing terms to avoid algebraic errors.