Differentiate the implicit functions: 𝑥^3 + 𝑦^2 = 6𝑥𝑦 + 3

To differentiate the implicit function \( x^3 + y^2 = 6xy + 3 \) with respect to \( x \), we’ll apply implicit differentiation, treating \( y \) as a function of \( x \) (i.e., \( y = y(x) \)).

### Steps:
1. Differentiate each term with respect to \( x \).
2. Apply the chain rule where necessary (for terms involving \( y \)), introducing \( \frac{dy}{dx} \) where \( y \) is differentiated.

#### Step-by-Step Solution:
1. **Differentiate the Left Side:**
   \[
   \frac{d}{dx}(x^3) + \frac{d}{dx}(y^2)
   \]
   - For \( x^3 \), we get \( 3x^2 \).
   - For \( y^2 \), use the chain rule: \( 2y \cdot \frac{dy}{dx} \).
   
   So, the left side becomes:
   \[
   3x^2 + 2y \frac{dy}{dx}
   \]

2. **Differentiate the Right Side:**
   \[
   \frac{d}{dx}(6xy) + \frac{d}{dx}(3)
   \]
   - For \( 6xy \), apply the product rule: 
     \[
     \frac{d}{dx}(6x) \cdot y + 6x \cdot \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}
     \]
   - For \( 3 \), since it is constant, \( \frac{d}{dx}(3) = 0 \).

   So, the right side becomes:
   \[
   6y + 6x \frac{dy}{dx}
   \]

3. **Combine Both Sides:**
   Now we have:
   \[
   3x^2 + 2y \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}
   \]

4. **Solve for \( \frac{dy}{dx} \):**
   Rearrange terms to isolate \( \frac{dy}{dx} \):
   \[
   2y \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2
   \]
   Factor out \( \frac{dy}{dx} \):
   \[
   (2y - 6x) \frac{dy}{dx} = 6y - 3x^2
   \]
   Divide by \( (2y - 6x) \) (assuming \( 2y - 6x \neq 0 \)):
   \[
   \frac{dy}{dx} = \frac{6y - 3x^2}{2y - 6x}
   \]

### Final Answer:
\[
\frac{dy}{dx} = \frac{6y - 3x^2}{2y - 6x}
\]

Would you like more details on any part of this differentiation process?

### Related Questions:
1. How do you solve implicit differentiation problems with higher powers of \( x \) and \( y \)?
2. What is the chain rule in differentiation, and how is it applied to implicit differentiation?
3. Can you apply implicit differentiation to trigonometric functions?
4. How can implicit differentiation be used to find the slope of a tangent line?
5. How do you know when to use implicit differentiation over explicit differentiation?

### Tip:
When differentiating implicitly, carefully apply the chain rule whenever \( y \) (or any other variable treated as a function of \( x \)) is differentiated.

Differentiate the implicit functions: x^3 + y^2 = 6xy + 3

Explore the implicit differentiation of x^3 + y^2 = 6xy + 3 with a complete, step-by-step solution. This guide covers essential calculus concepts like the chain rule and product rule, helping students understand how to find dy/dx in complex equations. Suitable for Grades 11-12 or early university-level calculus.

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