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Let's solve the problem.

The equation given is the equation of a circle:
\[
x^2 + y^2 = 100
\]
We need to find \( y' \) (the derivative of \( y \) with respect to \( x \)) at the point \( (6, 8) \).

### Step 1: Differentiate implicitly with respect to \( x \)
Differentiate both sides of the equation \( x^2 + y^2 = 100 \) with respect to \( x \):
\[
\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(100)
\]
On the left-hand side, use the chain rule:
\[
2x + 2y \frac{dy}{dx} = 0
\]
Simplify:
\[
2x + 2y y' = 0
\]
Divide through by 2:
\[
x + y y' = 0
\]

### Step 2: Solve for \( y' \)
Solve for \( y' \) (the derivative of \( y \) with respect to \( x \)):
\[
y y' = -x
\]
\[
y' = -\frac{x}{y}
\]

### Step 3: Substitute the point \( (6, 8) \)
Substitute \( x = 6 \) and \( y = 8 \) into the equation for \( y' \):
\[
y' = -\frac{6}{8} = -\frac{3}{4}
\]

So, \( y' = -\frac{3}{4} \).

The correct answer is:
\[
y' = -\frac{3}{4}
\]

Would you like more details or have any other questions?

### Related questions:
1. How do you apply implicit differentiation to other types of curves?
2. What is the geometric interpretation of the derivative \( y' \) on a circle?
3. How can you find the second derivative \( y'' \) in this case?
4. How would the result change if the equation represented an ellipse instead of a circle?
5. Can you use this technique to solve for \( y' \) in polar coordinates?

### Tip:
When differentiating implicitly, always remember to apply the chain rule to terms involving \( y \), since \( y \) is a function of \( x \).

Given the equation of the circle x^2 + y^2 = 100, find y' at the point (6, 8).

Learn how to find the derivative y' for the circle equation x^2 + y^2 = 100 at the point (6, 8). This problem involves implicit differentiation and applies the chain rule to solve for y'. Suitable for students in Grades 10-12.

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