We are given the equation:

$x^3(x + y) = y^2(2x - y)$

To find $\frac{dy}{dx}$ using implicit differentiation, we will differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate the left-hand side

The left-hand side is:

$x^3(x + y)$

We will apply the product rule: $\frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v'$.

Let $u = x^3$ and $v = (x + y)$.

• $u' = \frac{d}{dx}(x^3) = 3x^2$
• $v' = \frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}$ (since $\frac{d}{dx}(y) = \frac{dy}{dx}$)

Now applying the product rule:

$\frac{d}{dx}[x^3(x + y)] = 3x^2(x + y) + x^3(1 + \frac{dy}{dx})$

Step 2: Differentiate the right-hand side

The right-hand side is:

$y^2(2x - y)$

Again, we will use the product rule. Let $u = y^2$ and $v = (2x - y)$.

• $u' = \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$
• $v' = \frac{d}{dx}(2x - y) = 2 - \frac{dy}{dx}$

Now applying the product rule:

$\frac{d}{dx}[y^2(2x - y)] = 2y \cdot \frac{dy}{dx}(2x - y) + y^2(2 - \frac{dy}{dx})$

Step 3: Set up the equation

Now, equate the derivatives of both sides:

$3x^2(x + y) + x^3(1 + \frac{dy}{dx}) = 2y \cdot \frac{dy}{dx}(2x - y) + y^2(2 - \frac{dy}{dx})$

Step 4: Solve for $\frac{dy}{dx}$

Now, we need to solve this equation for $\frac{dy}{dx}$. First, group all terms involving $\frac{dy}{dx}$ on one side and all other terms on the other side.

Let me simplify and solve this equation.

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Would you like to see the full detailed solution for $\frac{dy}{dx}$? Or do you have any questions about this process?

Here are a few questions that could expand on this:

1. How can we verify the correctness of the implicit differentiation process?
2. What if the equation was in a different form (e.g., a rational equation)? How would that affect the differentiation?
3. Can you explain how to handle higher-order derivatives in implicit differentiation?
4. What are some common mistakes in implicit differentiation?
5. How would this process apply if $y$ were a function of multiple variables instead of just $x$?

Tip: When performing implicit differentiation, always remember to apply the chain rule whenever you differentiate terms involving $y$, since $y$ is treated as a function of $x$.