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We are given the equation \(\sqrt{x} + \sqrt{y} = 8\) and additional information \(y(4) = 36\). We need to find \(y'(4)\) by implicit differentiation.

### Step-by-Step Solution:

1. **Differentiate both sides implicitly** with respect to \(x\):
   \[
   \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(8)
   \]
   The right-hand side is 0 since the derivative of a constant is zero.

   Now, apply the chain rule to differentiate each square root term:
   \[
   \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}
   \]
   \[
   \frac{d}{dx}(\sqrt{y}) = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = \frac{1}{2\sqrt{y}} \cdot y'
   \]

   Thus, the differentiated equation becomes:
   \[
   \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot y' = 0
   \]

2. **Solve for \(y'\):**
   Rearrange the equation to isolate \(y'\):
   \[
   \frac{1}{2\sqrt{y}} \cdot y' = -\frac{1}{2\sqrt{x}}
   \]
   Multiply both sides by \(2\sqrt{y}\):
   \[
   y' = -\frac{\sqrt{y}}{\sqrt{x}}
   \]

3. **Substitute \(x = 4\) and \(y(4) = 36\):**
   We are given that when \(x = 4\), \(y = 36\). Substitute these values into the expression for \(y'\):
   \[
   y'(4) = -\frac{\sqrt{36}}{\sqrt{4}} = -\frac{6}{2} = -3
   \]

Thus, \(y'(4) = -3\).

---

Would you like further details or clarifications on any step? Here are some related questions to deepen understanding:

1. How is implicit differentiation different from explicit differentiation?
2. Why do we need to use the chain rule for the derivative of \(\sqrt{y}\)?
3. Can you apply implicit differentiation to an equation like \(x^2 + y^2 = 1\)?
4. What other types of functions require implicit differentiation?
5. How would the result change if \(y(x)\) were given in terms of \(x\) explicitly?

**Tip:** Always remember to apply the chain rule when differentiating a function of \(y\) with respect to \(x\) implicitly!

If sqrt(x) + sqrt(y) = 8 and y(4) = 36, find y'(4) by implicit differentiation.

This problem uses implicit differentiation to find y'(4) when sqrt(x) + sqrt(y) = 8 and y(4) = 36. Learn how to differentiate square roots and apply the chain rule. Suitable for students in Grades 11-12.

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