We are asked to differentiate the equation $x^3 + y^3 = 9xy$ implicitly to find $\frac{dy}{dx}$.

Step 1: Differentiate both sides with respect to $x$

The given equation is:

$x^3 + y^3 = 9xy$

We'll differentiate each term with respect to $x$. Remember that $y$ is a function of $x$, so we'll need to apply the chain rule for terms involving $y$.

Left-hand side:

• Differentiate $x^3$ with respect to $x$:
  $\frac{d}{dx}(x^3) = 3x^2$
• Differentiate $y^3$ with respect to $x$ using the chain rule:
  $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$

So, the derivative of the left-hand side is:

$3x^2 + 3y^2 \frac{dy}{dx}$

Right-hand side:

• Differentiate $9xy$ using the product rule:
  $\frac{d}{dx}(9xy) = 9 \left( x \frac{dy}{dx} + y \right)$

So, the derivative of the right-hand side is:

$9x \frac{dy}{dx} + 9y$

Step 2: Set up the equation for $\frac{dy}{dx}$

Now, equate the derivatives of both sides:

$3x^2 + 3y^2 \frac{dy}{dx} = 9x \frac{dy}{dx} + 9y$

Step 3: Collect terms involving $\frac{dy}{dx}$

Move all terms involving $\frac{dy}{dx}$ to one side of the equation:

$3y^2 \frac{dy}{dx} - 9x \frac{dy}{dx} = 9y - 3x^2$

Factor out $\frac{dy}{dx}$ on the left-hand side:

$\left( 3y^2 - 9x \right) \frac{dy}{dx} = 9y - 3x^2$

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

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x}$

Factor the numerator and denominator to simplify:

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

Cancel the common factor of 3:

$\frac{dy}{dx} = \frac{3y - x^2}{y^2 - 3x}$

Conclusion:

We have shown that:

$\frac{dy}{dx} = \frac{3y - x^2}{y^2 - 3x}$

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Would you like further clarification or a deeper breakdown of the steps?

Here are some related questions for you:

1. How would you solve for the second derivative $\frac{d^2y}{dx^2}$ for this curve?
2. What happens if we set $y = x$ in the equation $x^3 + y^3 = 9xy$?
3. How do implicit derivatives apply to more complicated equations involving trigonometric or exponential functions?
4. Can we use implicit differentiation for parametric equations? If so, how?
5. What is the relationship between implicit differentiation and the chain rule?

Tip: Always remember that when differentiating terms with $y$ in implicit differentiation, you must apply the chain rule since $y$ is treated as a function of $x$.