find derivative of: f(x)=1/sqrt(2x) using the definition of a derivative formula

To find the derivative of \( f(x) = \frac{1}{\sqrt{2x}} \) using the definition of a derivative, we'll use the following formula:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

### Step-by-Step Process:

1. **Substitute the function \( f(x) \) into the definition of the derivative:**
   \[
   f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2(x+h)}} - \frac{1}{\sqrt{2x}}}{h}
   \]

2. **Find a common denominator for the terms in the numerator:**

   The common denominator between \( \frac{1}{\sqrt{2(x+h)}} \) and \( \frac{1}{\sqrt{2x}} \) is \( \sqrt{2(x+h)} \cdot \sqrt{2x} \).

   So, the expression becomes:
   \[
   f'(x) = \lim_{h \to 0} \frac{\frac{\sqrt{2x} - \sqrt{2(x+h)}}{\sqrt{2(x+h)} \cdot \sqrt{2x}}}{h}
   \]

3. **Simplify the expression:**

   This simplifies to:
   \[
   f'(x) = \lim_{h \to 0} \frac{\sqrt{2x} - \sqrt{2(x+h)}}{h \cdot \sqrt{2(x+h)} \cdot \sqrt{2x}}
   \]

4. **Use the difference of squares to simplify further:**

   Multiply both the numerator and the denominator by \( \sqrt{2x} + \sqrt{2(x+h)} \), using the identity \( a^2 - b^2 = (a - b)(a + b) \):

   \[
   f'(x) = \lim_{h \to 0} \frac{(2x - 2(x+h))}{h \cdot \sqrt{2(x+h)} \cdot \sqrt{2x} \cdot (\sqrt{2x} + \sqrt{2(x+h)})}
   \]

   Simplifying the numerator:
   \[
   f'(x) = \lim_{h \to 0} \frac{(2x - 2x - 2h)}{h \cdot \sqrt{2(x+h)} \cdot \sqrt{2x} \cdot (\sqrt{2x} + \sqrt{2(x+h)})}
   \]

   \[
   f'(x) = \lim_{h \to 0} \frac{-2h}{h \cdot \sqrt{2(x+h)} \cdot \sqrt{2x} \cdot (\sqrt{2x} + \sqrt{2(x+h)})}
   \]

5. **Cancel out \( h \) in the numerator and denominator:**

   \[
   f'(x) = \lim_{h \to 0} \frac{-2}{\sqrt{2(x+h)} \cdot \sqrt{2x} \cdot (\sqrt{2x} + \sqrt{2(x+h)})}
   \]

6. **Take the limit as \( h \to 0 \):**

   As \( h \to 0 \), \( \sqrt{2(x+h)} \) approaches \( \sqrt{2x} \). Therefore, the expression becomes:

   \[
   f'(x) = \frac{-2}{\sqrt{2x} \cdot \sqrt{2x} \cdot (2\sqrt{2x})}
   \]

7. **Simplify the final expression:**

   \[
   f'(x) = \frac{-2}{2x \cdot 2\sqrt{2x}} = \frac{-1}{2x^{3/2}}
   \]

Thus, the derivative of \( f(x) = \frac{1}{\sqrt{2x}} \) is:

\[
f'(x) = \frac{-1}{2x^{3/2}}
\]

Would you like to go into more details or have any further questions?

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Here are 5 related questions:
1. How would you find the second derivative of \( f(x) = \frac{1}{\sqrt{2x}} \)?
2. What happens when we apply the chain rule instead of the definition of the derivative?
3. How do you differentiate \( f(x) = \frac{1}{x^n} \) for any \( n \)?
4. Can you explain the geometric interpretation of the derivative in this case?
5. What is the derivative of \( f(x) = \frac{1}{\sqrt{x}} \), and how does it compare?

**Tip:** When simplifying complex expressions, always check for algebraic identities like difference of squares or common denominators!

Explore the step-by-step process to find the derivative of f(x) = 1/sqrt(2x) using the definition of a derivative. Learn key calculus concepts, including limits, the difference of squares, and algebraic simplification. This problem is ideal for advanced calculus students in Grades 11-12.

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