Differentiate from first principles:
4. `(1)/( sqrtx)`

Answer

To differentiate the function \( f(x) = \frac{1}{\sqrt{x}} \) from first principles, we will follow these steps:

### Step 1: Define the function and the difference quotient
We start with the function:
\[ f(x) = \frac{1}{\sqrt{x}} \]
According to the first principle of differentiation, the derivative \( f'(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x+h) \) and \( f(x) \)
Now, we need to find \( f(x+h) \):
\[ f(x+h) = \frac{1}{\sqrt{x+h}} \]
Substituting 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 for the fractions in the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x} \cdot \sqrt{x+h} \cdot h} \]

### Step 4: Rationalize the numerator
Next, we rationalize the numerator by multiplying by the conjugate:
\[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{(\sqrt{x} + \sqrt{x+h}) \cdot \sqrt{x} \cdot \sqrt{x+h} \cdot h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{x - (x+h)}{(\sqrt{x} + \sqrt{x+h}) \cdot \sqrt{x} \cdot \sqrt{x+h} \cdot h} \]
\[ = \lim_{h \to 0} \frac{-h}{(\sqrt{x} + \sqrt{x+h}) \cdot \sqrt{x} \cdot \sqrt{x+h} \cdot h} \]

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

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

### Final Result
Thus, the derivative of \( f(x) = \frac{1}{\sqrt{x}} \) is:
\[ f'(x) = -\frac{1}{2x^{3/2}} \]