implicit\:derivative\:\frac{dy}{dx},\:x^2-3y^2=2xy^3

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 \(x\).

implicit derivative: \frac{dy}{dx}, x^2 - 3y^2 = 2xy^3

Find the implicit derivative \frac{dy}{dx} of the equation x^2 - 3y^2 = 2xy^3 with a detailed step-by-step solution. This problem involves applying implicit differentiation, the product rule, and the chain rule. Ideal for students in Grades 11-12 or college-level calculus.

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