Let sin(y2 − 4x) = 0. and w = ln(3x2 −2y); x = t1 2 ; y = t2 3  Find dy dx for the first function.

To find \( \frac{dy}{dx} \) for the given first function \( \sin(y^2 - 4x) = 0 \), we will use **implicit differentiation**.

### Step-by-Step Solution:

1. **Given equation:**
   \[
   \sin(y^2 - 4x) = 0
   \]
   
   This implies that:
   \[
   y^2 - 4x = n\pi
   \]
   where \( n \in \mathbb{Z} \) because \( \sin(n\pi) = 0 \).

2. **Differentiate both sides with respect to \(x\):**
   Apply the chain rule to both sides of the equation:
   \[
   \frac{d}{dx}(y^2 - 4x) = \frac{d}{dx}(n\pi)
   \]
   
   Since \(n\pi\) is a constant, its derivative is zero. So, we get:
   \[
   2y \frac{dy}{dx} - 4 = 0
   \]

3. **Solve for \( \frac{dy}{dx} \):**
   Rearrange the equation to isolate \( \frac{dy}{dx} \):
   \[
   2y \frac{dy}{dx} = 4
   \]
   \[
   \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}
   \]

Thus, the derivative of \(y\) with respect to \(x\) is:
\[
\boxed{\frac{dy}{dx} = \frac{2}{y}}
\]

Would you like details or have any questions on this?

### Related Questions:
1. How do we differentiate implicitly when dealing with trigonometric functions?
2. What happens if \( \sin(\theta) = 0 \) but the argument \( \theta \) involves higher powers of \(x\)?
3. Can you explain how the chain rule works in implicit differentiation?
4. Why do we need the constant \( n\pi \) in the equation?
5. What is the geometric meaning of \( \frac{dy}{dx} \) for this implicit function?

#### Tip:
When dealing with trigonometric functions in implicit differentiation, always account for periodicity and constants like \(n\pi\) for solving the equations.

Let sin(y^2 − 4x) = 0. and w = ln(3x^2 − 2y); x = t1^2 ; y = t2^3. Find dy/dx for the first function.

Explore the implicit differentiation of the trigonometric equation sin(y^2 - 4x) = 0. This problem uses the chain rule and involves finding dy/dx for advanced calculus students. Suitable for Grades 11-12.

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