find the derivative of  1/square root of x by using the definition of the derivative

To find the derivative of \( f(x) = \frac{1}{\sqrt{x}} \) using the definition of the derivative, we start with the standard formula for the derivative:

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

### Step-by-Step Process:

1. **Set up the function**: The given function is \( f(x) = \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \).

2. **Substitute into the definition**:
   Using the definition of the derivative:

   \[
   f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}
   \]

3. **Combine the two terms in the numerator**:
   To combine the terms, we need to get a common denominator in the numerator:

   \[
   \frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}} = \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x} \cdot \sqrt{x+h}}
   \]

   So, the expression becomes:

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

   Simplifying:

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

4. **Multiply by the conjugate**:
   To simplify \( \sqrt{x} - \sqrt{x+h} \), multiply the numerator and denominator by the conjugate \( \sqrt{x} + \sqrt{x+h} \):

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

   Using the difference of squares formula in the numerator:

   \[
   (\sqrt{x})^2 - (\sqrt{x+h})^2 = x - (x+h) = -h
   \]

   So, the expression simplifies to:

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

5. **Cancel the \( h \) terms**:
   The \( h \) in the numerator and denominator cancels:

   \[
   f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x} \cdot \sqrt{x+h} \cdot (\sqrt{x} + \sqrt{x+h})}
   \]

6. **Evaluate the limit as \( h \to 0 \)**:
   As \( h \to 0 \), \( \sqrt{x+h} \to \sqrt{x} \), so the expression becomes:

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

   Simplifying:

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

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

This is the derivative of \( f(x) = \frac{1}{\sqrt{x}} \) using the definition of the derivative.

Would you like further explanation or more details?

### Related Questions:
1. How would you find the derivative of \( \frac{1}{x^2} \) using the definition of the derivative?
2. What is the interpretation of a derivative in terms of rates of change?
3. How can you use the power rule to verify the derivative of \( x^{-\frac{1}{2}} \)?
4. How does multiplying by a conjugate help simplify radical expressions?
5. Can you apply the definition of the derivative to a trigonometric function like \( \sin(x) \)?

### Tip:
Always simplify complex fractions by multiplying by the conjugate when dealing with square roots—it helps eliminate the radicals in the numerator.

find the derivative of 1/square root of x by using the definition of the derivative

This solution explains how to find the derivative of the function 1/√x step-by-step using the definition of the derivative. The process involves limits, simplifying radical expressions, and applying the difference of squares. Suitable for high school students or anyone studying introductory calculus.

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