Differentiate from first principles:
9. `sqrt( 2x +3)`

Answer

To differentiate the function \( f(x) = \sqrt{2x + 3} \) from first principles, we follow these steps:

### Step 1: Define the function
Let \( f(x) = \sqrt{2x + 3} \).

### Step 2: Use the definition of the derivative
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 3: Substitute \( f(x + h) \) and \( f(x) \)
Now we need to find \( f(x + h) \):
\[ f(x + h) = \sqrt{2(x + h) + 3} = \sqrt{2x + 2h + 3} \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{2x + 2h + 3} - \sqrt{2x + 3}}{h} \]

### Step 4: Rationalize the numerator
To simplify the expression, we multiply the numerator and denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{2x + 2h + 3} - \sqrt{2x + 3}\right) \left(\sqrt{2x + 2h + 3} + \sqrt{2x + 3}\right)}{h \left(\sqrt{2x + 2h + 3} + \sqrt{2x + 3}\right)} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{(2x + 2h + 3) - (2x + 3)}{h \left(\sqrt{2x + 2h + 3} + \sqrt{2x + 3}\right)} \]

### Step 5: Simplify the expression
The numerator simplifies as follows:
\[ (2x + 2h + 3) - (2x + 3) = 2h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{2h}{h \left(\sqrt{2x + 2h + 3} + \sqrt{2x + 3}\right)} \]
Now we can cancel \( h \) in the numerator and denominator:
\[ f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 3} + \sqrt{2x + 3}} \]

### Step 6: Evaluate the limit
As \( h \to 0 \), \( \sqrt{2x + 2h + 3} \) approaches \( \sqrt{2x + 3} \):
\[ f'(x) = \frac{2}{\sqrt{2x + 3} + \sqrt{2x + 3}} = \frac{2}{2\sqrt{2x + 3}} = \frac{1}{\sqrt{2x + 3}} \]

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