To find \( y' \) (the derivative of \( y \) with respect to \( x \)), we'll implicitly differentiate the equation \( 16x^2 - y^2 = 1 \) with respect to \( x \).

Given:
\[
16x^2 - y^2 = 1
\]

### Step 1: Differentiate both sides with respect to \( x \)

\[
\frac{d}{dx}(16x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}(1)
\]

### Step 2: Apply the derivatives

- The derivative of \( 16x^2 \) with respect to \( x \) is \( 32x \).
- The derivative of \( y^2 \) with respect to \( x \) is \( 2y \cdot y' \) (using the chain rule).
- The derivative of the constant \( 1 \) is \( 0 \).

So, differentiating gives:
\[
32x - 2yy' = 0
\]

### Step 3: Solve for \( y' \)

Move the \( 2yy' \) term to the other side:
\[
32x = 2yy'
\]

Now, solve for \( y' \):
\[
y' = \frac{32x}{2y} = \frac{16x}{y}
\]

So the derivative \( y' \) is:
\[
y' = \frac{16x}{y}
\]

Would you like further details or have any questions?

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Here are 5 related questions:
1. How would you apply the chain rule to more complex functions involving products of terms?
2. What happens if the equation was \( 16x^2 + y^2 = 1 \)? How would that affect \( y' \)?
3. How can implicit differentiation be applied to trigonometric identities?
4. What are the advantages of using implicit differentiation over solving for \( y \) first?
5. How would you find the second derivative \( y'' \) from this equation?

**Tip:** When using implicit differentiation, carefully apply the chain rule, especially when dealing with terms like \( y^2 \) where \( y \) is a function of \( x \).

Learn how to find the derivative y' (y prime) from the equation 16x² - y² = 1 using implicit differentiation. This detailed step-by-step solution covers implicit differentiation, derivatives, and the chain rule. Suitable for Grades 11-12 students.

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