We are given the equation $x^4 y - xy^3 = -2$ and asked to find the equation of the tangent line to the graph at the point $(-1, -1)$ using implicit differentiation.

Step-by-step Solution:

1. Differentiate the equation implicitly with respect to $x$: $\frac{d}{dx} \left( x^4 y - xy^3 \right) = \frac{d}{dx}(-2)$ For each term:

   • The derivative of $x^4 y$ using the product rule: = 4x^3 y + x^4 \frac{dy}{dx}$$
   • The derivative of $xy^3$ also using the product rule: = y^3 + x \cdot 3y^2 \frac{dy}{dx}$$
   • The derivative of the constant $-2$ is 0.

2. Substitute these into the equation: $4x^3 y + x^4 \frac{dy}{dx} - (y^3 + 3xy^2 \frac{dy}{dx}) = 0$ Simplifying: $4x^3 y + x^4 \frac{dy}{dx} - y^3 - 3xy^2 \frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$: Combine the terms with $\frac{dy}{dx}$: $x^4 \frac{dy}{dx} - 3xy^2 \frac{dy}{dx} = y^3 - 4x^3 y$ Factor out $\frac{dy}{dx}$ on the left-hand side: $\frac{dy}{dx}(x^4 - 3xy^2) = y^3 - 4x^3 y$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y^3 - 4x^3 y}{x^4 - 3xy^2}$

4. Substitute the point $(-1, -1)$ into the derivative:

   • $x = -1$
   • $y = -1$

   Substituting these values into the expression for $\frac{dy}{dx}$:

   = \frac{-1 - 4}{1 - 3} = \frac{-5}{-2} = \frac{5}{2}$$

5. Find the equation of the tangent line: The slope of the tangent line at $(-1, -1)$ is $\frac{5}{2}$. The equation of the tangent line is given by: $y - y_1 = m(x - x_1)$ Substituting $m = \frac{5}{2}, x_1 = -1, y_1 = -1$: $y + 1 = \frac{5}{2}(x + 1)$ Simplifying: $y + 1 = \frac{5}{2}x + \frac{5}{2}$ $y = \frac{5}{2}x + \frac{5}{2} - 1$ $y = \frac{5}{2}x + \frac{3}{2}$

Thus, the equation of the tangent line is: $y = \frac{5}{2}x + \frac{3}{2}$

Would you like further details on any step? Here are some related questions for further exploration:

1. How can we apply the product rule to more complex functions?
2. What is the significance of implicit differentiation in calculus?
3. How does one generalize the method to find higher-order derivatives in implicit differentiation?
4. How do we interpret the slope of the tangent line geometrically?
5. What are other methods of finding tangent lines to curves besides implicit differentiation?

Tip: Always remember to use the product rule when differentiating terms that are a product of two functions, especially in implicit differentiation.