To find the implicit derivative $\frac{dy}{dx}$ of the equation $x^2 - 3y^2 = 2xy^3$, we differentiate both sides of the equation with respect to $x$, applying the rules of implicit differentiation (remembering that $y$ is a function of $x$).

The equation to differentiate is: $x^2 - 3y^2 = 2xy^3$

Step 1: Differentiate the left-hand side

• For $x^2$, the derivative with respect to $x$ is: $\frac{d}{dx}(x^2) = 2x$
• For $-3y^2$, apply the chain rule (since $y$ is a function of $x$): $\frac{d}{dx}(-3y^2) = -6y \frac{dy}{dx}$

Thus, the derivative of the left-hand side is: $2x - 6y \frac{dy}{dx}$

Step 2: Differentiate the right-hand side

For $2xy^3$, we apply the product rule $\frac{d}{dx}(uv) = u'v + uv'$:

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

Applying the product rule gives: $\frac{d}{dx}(2xy^3) = 2y^3 + 6xy^2 \frac{dy}{dx}$

Step 3: Put it all together

Now substitute the derivatives back into the equation: $2x - 6y \frac{dy}{dx} = 2y^3 + 6xy^2 \frac{dy}{dx}$

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

Move all terms involving $\frac{dy}{dx}$ to one side and the rest to the other side: $-6y \frac{dy}{dx} - 6xy^2 \frac{dy}{dx} = 2y^3 - 2x$ Factor out $\frac{dy}{dx}$ on the left-hand side: $\frac{dy}{dx}(-6y - 6xy^2) = 2y^3 - 2x$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2y^3 - 2x}{-6y - 6xy^2}$ Simplifying: $\frac{dy}{dx} = \frac{y^3 - x}{3y(1 + xy)}$

This is the implicit derivative $\frac{dy}{dx}$.

Would you like further explanation on any of the steps?

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Here are five related questions you might explore:

1. What is the derivative of $x^2 + y^2 = r^2$ using implicit differentiation?
2. How does the chain rule apply in implicit differentiation?
3. Can implicit differentiation be used for higher-order derivatives?
4. What is the geometric interpretation of implicit differentiation?
5. How would the derivative change if we introduced a new variable in the original equation?

Tip: In implicit differentiation, always apply the chain rule when differentiating terms involving $y$, since $y$ is considered a function of $x$.