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!