Find dy for the given values of x and Upper Delta x.
yequalsStartRoot 2 x plus 6 EndRoot​; xequals1​; Upper Delta xequals0.12
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Part 1
What is the derivative of yequalsStartRoot 2 x plus 6 EndRoot​?

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \sqrt{2x + 6} \), we can start by applying the chain rule. Here's a step-by-step solution.

1. **Rewrite the function for clarity**:
   \[
   y = (2x + 6)^{\frac{1}{2}}
   \]

2. **Differentiate with respect to \( x \)**:
   Using the chain rule, we get:
   \[
   \frac{dy}{dx} = \frac{1}{2}(2x + 6)^{-\frac{1}{2}} \cdot \frac{d}{dx}(2x + 6)
   \]

3. **Calculate the derivative inside**:
   \[
   \frac{d}{dx}(2x + 6) = 2
   \]

4. **Combine terms**:
   \[
   \frac{dy}{dx} = \frac{1}{2} \cdot (2x + 6)^{-\frac{1}{2}} \cdot 2
   \]

5. **Simplify**:
   \[
   \frac{dy}{dx} = \frac{1}{\sqrt{2x + 6}}
   \]

So, the derivative of \( y = \sqrt{2x + 6} \) is:
\[
\frac{dy}{dx} = \frac{1}{\sqrt{2x + 6}}
\]

Let me know if you’d like to go further and find \( dy \) for \( x = 1 \) and \( \Delta x = 0.12 \). 

---

Would you like to continue?

1. How do you apply the chain rule to different functions?
2. How does simplifying radicals help in derivative calculations?
3. Why is the chain rule useful for nested functions?
4. What does \( \Delta x \) represent in this context?
5. How is the concept of \( dy \) or differential applied in real-world problems?

**Tip:** Always simplify expressions as much as possible to make differentiation easier!

Find dy for the given values of x and Δx. y = √(2x + 6); x = 1; Δx = 0.12. What is the derivative of y = √(2x + 6)?

Learn how to find the differential dy for y = √(2x + 6) given x = 1 and Δx = 0.12. This step-by-step guide covers the chain rule and differential applications, ideal for Grades 11-12 calculus students.

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