We are tasked with finding $\frac{dy}{dx}$ using implicit differentiation for the equation:

$4x^3 + x^2y - xy^3 = 2$

Step-by-Step Implicit Differentiation:

1. Differentiate each term with respect to $x$ while treating $y$ as a function of $x$ (i.e., apply the chain rule when differentiating terms involving $y$).

First term: $4x^3$

$\frac{d}{dx}[4x^3] = 12x^2$

Second term: $x^2y$

We use the product rule for $x^2y$, where $x^2$ is one function and $y$ is the other: [ \frac{d}{dx}[x^2y] = \frac{d}{dx}[x^2] \cdot y + x^2 \cdot \frac{dy}{dx} = 2xy + x^2\frac{dy}{dx} ]

Third term: $-xy^3$

Again, apply the product rule for $xy^3$, where $x$ is one function and $y^3$ is the other: [ \frac{d}{dx}[-xy^3] = -\left(\frac{d}{dx}[x] \cdot y^3 + x \cdot \frac{d}{dx}[y^3]\right) = -y^3 - 3xy^2\frac{dy}{dx} ]

Differentiate the right-hand side:

$\frac{d}{dx}[2] = 0$

Combine all the differentiated parts:

$12x^2 + 2xy + x^2\frac{dy}{dx} - y^3 - 3xy^2\frac{dy}{dx} = 0$

Isolate $\frac{dy}{dx}$:

Group the terms involving $\frac{dy}{dx}$ on one side: $x^2\frac{dy}{dx} - 3xy^2\frac{dy}{dx} = y^3 - 12x^2 - 2xy$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x^2 - 3xy^2) = y^3 - 12x^2 - 2xy$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y^3 - 12x^2 - 2xy}{x^2 - 3xy^2}$

Thus, the implicit derivative is: $\frac{dy}{dx} = \frac{y^3 - 12x^2 - 2xy}{x^2 - 3xy^2}$

Would you like a detailed explanation of any step, or do you have any questions?

Here are 5 questions that expand on this topic:

1. How does implicit differentiation differ from explicit differentiation?
2. What is the product rule, and why is it used here?
3. How would the result change if there were higher powers of $y$ involved?
4. Can you apply implicit differentiation to trigonometric functions?
5. How does this process extend to higher-order derivatives?

Tip: When applying the product rule in implicit differentiation, always remember to differentiate $y$ as a function of $x$, which adds a $\frac{dy}{dx}$ term.