Find 𝑑𝑦𝑑𝑥
 by implicit differentiation.

2𝑥3𝑦2−4𝑥2𝑦=4

To find \(\frac{dy}{dx}\) by implicit differentiation for the given equation:

\[
2x^3y^2 - 4x^2y = 4
\]

We differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\) (i.e., applying the chain rule whenever we differentiate \(y\)).

### Step 1: Differentiate the left-hand side

Start with the first term: \( 2x^3y^2 \)
- Differentiate using the product rule: \(\frac{d}{dx}(u \cdot v) = u'v + uv'\), where \(u = 2x^3\) and \(v = y^2\).
  \[
  \frac{d}{dx}(2x^3y^2) = \frac{d}{dx}(2x^3) \cdot y^2 + 2x^3 \cdot \frac{d}{dx}(y^2)
  \]
  - \(\frac{d}{dx}(2x^3) = 6x^2\)
  - \(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\) (by chain rule)

  Thus,
  \[
  \frac{d}{dx}(2x^3y^2) = 6x^2y^2 + 4x^3y \cdot \frac{dy}{dx}
  \]

Next, differentiate the second term: \( -4x^2y \)
- Again, apply the product rule: \(\frac{d}{dx}(u \cdot v) = u'v + uv'\), where \(u = -4x^2\) and \(v = y\).
  \[
  \frac{d}{dx}(-4x^2y) = \frac{d}{dx}(-4x^2) \cdot y + (-4x^2) \cdot \frac{d}{dx}(y)
  \]
  - \(\frac{d}{dx}(-4x^2) = -8x\)
  - \(\frac{d}{dx}(y) = \frac{dy}{dx}\)

  Thus,
  \[
  \frac{d}{dx}(-4x^2y) = -8xy - 4x^2 \cdot \frac{dy}{dx}
  \]

### Step 2: Differentiate the right-hand side

Since the right-hand side is just a constant (4), its derivative is 0:
\[
\frac{d}{dx}(4) = 0
\]

### Step 3: Combine the results

Now, combine the differentiated terms:
\[
6x^2y^2 + 4x^3y \cdot \frac{dy}{dx} - 8xy - 4x^2 \cdot \frac{dy}{dx} = 0
\]

### Step 4: Collect terms involving \(\frac{dy}{dx}\)

Group the terms involving \(\frac{dy}{dx}\):
\[
4x^3y \cdot \frac{dy}{dx} - 4x^2 \cdot \frac{dy}{dx} = -6x^2y^2 + 8xy
\]

### Step 5: Factor out \(\frac{dy}{dx}\)

Factor out \(\frac{dy}{dx}\) on the left-hand side:
\[
\frac{dy}{dx}(4x^3y - 4x^2) = -6x^2y^2 + 8xy
\]

### Step 6: Solve for \(\frac{dy}{dx}\)

Finally, solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{-6x^2y^2 + 8xy}{4x^3y - 4x^2}
\]

We can simplify this expression by factoring the numerator and denominator if necessary.

Let me know if you want to explore further simplifications or if you have any questions!

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Here are five related questions to deepen your understanding:

1. What are the steps of the product rule, and how is it applied in implicit differentiation?
2. How does the chain rule affect differentiation when dealing with products of functions of \(y\)?
3. Can you simplify the final expression for \(\frac{dy}{dx}\) further by factoring?
4. How would you approach implicit differentiation if there were higher powers of \(y\)?
5. What are the main differences between explicit and implicit differentiation?

**Tip:** When dealing with implicit differentiation, always remember to apply the chain rule carefully when differentiating terms involving \(y\).

Find \(\frac{dy}{dx}\) by implicit differentiation. 2x^3y^2 - 4x^2y = 4

Discover how to find \(\frac{dy}{dx}\) using implicit differentiation for the equation 2x^3y^2 - 4x^2y = 4. This problem explores key calculus concepts, including the product rule and chain rule, suitable for high school students in Grades 10-12.

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