`1/(sqrt(2x + 3))`

Answer

To differentiate the function \( f(x) = \frac{1}{\sqrt{2x + 3}} \) using the first principles of differentiation, we will follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) using the first principles is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) and \( f(x+h) \)
We have:
\[ f(x) = \frac{1}{\sqrt{2x + 3}} \]
Now, we need to find \( f(x+h) \):
\[ f(x+h) = \frac{1}{\sqrt{2(x+h) + 3}} = \frac{1}{\sqrt{2x + 2h + 3}} \]

### Step 3: Set up the limit
Now we can substitute \( f(x) \) and \( f(x+h) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2x + 2h + 3}} - \frac{1}{\sqrt{2x + 3}}}{h} \]

### Step 4: Simplify the expression
To simplify the expression, we will combine the fractions in the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{2x + 3} - \sqrt{2x + 2h + 3}}{h \cdot \sqrt{2x + 3} \cdot \sqrt{2x + 2h + 3}} \]

### Step 5: Rationalize the numerator
To eliminate the square roots in the numerator, we multiply by the conjugate:
\[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{2x + 3} - \sqrt{2x + 2h + 3})(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})}{h \cdot \sqrt{2x + 3} \cdot \sqrt{2x + 2h + 3} \cdot (\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{(2x + 3) - (2x + 2h + 3)}{h \cdot \sqrt{2x + 3} \cdot \sqrt{2x + 2h + 3} \cdot (\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \]

### Step 6: Simplify the numerator further
The numerator simplifies to:
\[ 2x + 3 - 2x - 2h - 3 = -2h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{-2h}{h \cdot \sqrt{2x + 3} \cdot \sqrt{2x + 2h + 3} \cdot (\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \]

### Step 7: Cancel \( h \) and evaluate the limit
Cancel \( h \) from the numerator and denominator:
\[ f'(x) = \lim_{h \to 0} \frac{-2}{\sqrt{2x + 3} \cdot \sqrt{2x + 2h + 3} \cdot (\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \]
As \( h \to 0 \), \( \sqrt{2x + 2h + 3} \to \sqrt{2x + 3} \):
\[ f'(x) = \frac{-2}{\sqrt{2x + 3} \cdot \sqrt{2x + 3} \cdot (2\sqrt{2x + 3})} = \frac{-2}{2(2x + 3)^{3/2}} = \frac{-1}{(2x + 3)^{3/2}} \]

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