Differentiate `sqrt(x)` with respect to `x` from definition.

Answer

To differentiate \( f(x) = \sqrt{x} \) with respect to \( x \) using the first principle of differentiation, we follow these steps:

### Step 1: Define the function and the increment
Let \( f(x) = \sqrt{x} \). According to the first principle of differentiation, we need to find \( f'(x) \) using the limit definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the limit
Substituting \( f(x) \) into the limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \]

### Step 3: Rationalize the numerator
To simplify the expression, we multiply the numerator and the denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \]
This gives:
\[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} \]

### Step 4: Simplify the numerator
Using the difference of squares, the numerator simplifies to:
\[ (\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x = h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} \]

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

### Step 6: Apply the limit
Now, we apply the limit as \( h \) approaches 0:
\[ f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}} \]

### Final Answer
Thus, the derivative of \( \sqrt{x} \) with respect to \( x \) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]

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