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

$x^3 + y^3 = 18xy$

Step-by-step solution:

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

$\frac{d}{dx}\left(x^3 + y^3\right) = \frac{d}{dx}\left(18xy\right)$

2. Apply the chain rule for $y^3$ (since $y$ is a function of $x$), and the product rule for $18xy$:

   • Derivative of $x^3$ is $3x^2$.
   • Derivative of $y^3$ is $3y^2 \frac{dy}{dx}$ (using the chain rule).
   • Derivative of $18xy$ (using the product rule): $\frac{d}{dx}\left(18xy\right) = 18x \frac{dy}{dx} + 18y$

Now, the differentiated equation is:

$3x^2 + 3y^2 \frac{dy}{dx} = 18x \frac{dy}{dx} + 18y$

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

Rearrange the equation to group all terms with $\frac{dy}{dx}$ on one side:

$3y^2 \frac{dy}{dx} - 18x \frac{dy}{dx} = 18y - 3x^2$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx} \left( 3y^2 - 18x \right) = 18y - 3x^2$

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

$\frac{dy}{dx} = \frac{18y - 3x^2}{3y^2 - 18x}$

Simplify the expression:

$\frac{dy}{dx} = \frac{6(3y - x^2)}{3(y^2 - 6x)}$

Thus, the derivative is:

$\frac{dy}{dx} = \frac{6(3y - x^2)}{3(y^2 - 6x)}$

Would you like to see more details or ask any clarifying questions? Here are 5 related questions for practice:

1. How do you apply implicit differentiation to trigonometric functions?
2. Can you find $\frac{dy}{dx}$ for $x^2 + y^2 = r^2$ using implicit differentiation?
3. What is the geometric interpretation of implicit differentiation?
4. How does implicit differentiation work when both variables are squared?
5. What is the product rule, and how do you apply it in implicit differentiation?

Tip:

When using implicit differentiation, remember to apply the chain rule whenever you're differentiating terms involving $y$, as $y$ is treated as a function of $x$.