Given `f(x)=1/sqrt(x), x gt 0`, find `f^(')(x)` by delta method.

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = \frac{1}{\sqrt{x}} \) using the delta method, we will follow these steps:

### Step 1: Define the derivative using the limit definition
The derivative \( f'(x) \) can be defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the limit
We know that:
\[ f(x) = \frac{1}{\sqrt{x}} \]
Now, we need to find \( f(x+h) \):
\[ f(x+h) = \frac{1}{\sqrt{x+h}} \]
Substituting this into the limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \]

### Step 3: Simplify the expression
To simplify the expression, we need a common denominator:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h} \cdot h} \]

### Step 4: Rationalize the numerator
To eliminate the square roots in the numerator, we multiply the numerator and the denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{h \cdot \sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{x - (x+h)}{h \cdot \sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \]
\[ = \lim_{h \to 0} \frac{-h}{h \cdot \sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \]

### Step 5: Cancel \( h \) in the numerator and denominator
We can cancel \( h \) from the numerator and denominator:
\[ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \]

### Step 6: Evaluate the limit as \( h \to 0 \)
As \( h \) approaches 0, \( \sqrt{x+h} \) approaches \( \sqrt{x} \):
\[ f'(x) = \frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x} + \sqrt{x})} = \frac{-1}{x \cdot 2\sqrt{x}} = \frac{-1}{2x\sqrt{x}} \]

### Final Result
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = -\frac{1}{2x\sqrt{x}} \]