To find $\frac{dy}{dx}$ (the derivative of $y$ with respect to $x$) for the equation:

$x^3y + xy^3 = 8$

we'll use implicit differentiation. Let's go through the steps:

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

We'll apply the product rule and chain rule where necessary.

$\frac{d}{dx}(x^3y) + \frac{d}{dx}(xy^3) = \frac{d}{dx}(8)$

Step 2: Differentiate each term

• For $x^3y$, we apply the product rule: $\frac{d}{dx}(x^3y) = \frac{d}{dx}(x^3) \cdot y + x^3 \cdot \frac{dy}{dx} = 3x^2 y + x^3 \frac{dy}{dx}$

• For $xy^3$, we again apply the product rule:

= y^3 + x \cdot 3y^2 \frac{dy}{dx}$$ The derivative of the constant 8 is zero: $$\frac{d}{dx}(8) = 0$$ ### Step 3: Put the pieces together Now we substitute the derivatives into the equation: $$3x^2 y + x^3 \frac{dy}{dx} + y^3 + 3xy^2 \frac{dy}{dx} = 0$$ ### Step 4: Solve for $$\frac{dy}{dx}$$ Group the terms involving $$\frac{dy}{dx}$$: $$x^3 \frac{dy}{dx} + 3xy^2 \frac{dy}{dx} = -3x^2 y - y^3$$ Factor out $$\frac{dy}{dx}$$: $$\frac{dy}{dx}(x^3 + 3xy^2) = -3x^2 y - y^3$$ Finally, solve for $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = \frac{-3x^2 y - y^3}{x^3 + 3xy^2}$$ This is the derivative of $$y$$ with respect to $$x$$. --- Would you like more details or further explanation? Here are some related questions to deepen your understanding: 1. What is implicit differentiation, and how is it applied in calculus? 2. How does the product rule work, and when is it necessary? 3. Why do we need to use the chain rule in this example? 4. Can you explain how to differentiate a function like $$y^3$$ with respect to $$x$$? 5. What are some common applications of implicit differentiation in physics? **Tip**: Always check if the equation you're differentiating implicitly has terms with both $$x$$ and $$y$$ multiplied together, as this indicates the product rule is needed.