Find, from first principles, the derivative of the following w.r.t. x :
`sqrt(x)`

Answer

To find the derivative of the function \( f(x) = \sqrt{x} \) using first principles, we will follow these steps:

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \sqrt{x} \]

2. **Use the definition of the derivative**:
The derivative of \( f(x) \) at a point \( x \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

3. **Substitute the function into the limit**:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \]

4. **Multiply and divide by the conjugate**:
To simplify the expression, multiply the numerator and denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} \]

5. **Apply the difference of squares**:
The numerator simplifies using the identity \( a^2 - b^2 = (a-b)(a+b) \):
\[ f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \]
\[ = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} \]

6. **Cancel \( h \) in the numerator and denominator**:
\[ = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} \]

7. **Evaluate the limit as \( h \to 0 \)**:
Substituting \( h = 0 \):
\[ = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}} \]

8. **Final result**:
Thus, the derivative of \( f(x) = \sqrt{x} \) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]