Find the derivative of `sqrt(3x + 5)` using first principle of derivative

Answer

To find the derivative of \( f(x) = \sqrt{3x + 5} \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the definition of the derivative using the first principle.
The first principle of derivatives states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Substitute the function into the formula.
Here, \( f(x) = \sqrt{3x + 5} \). Therefore, we have:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{3(x + h) + 5} - \sqrt{3x + 5}}{h} \]

### Step 3: Simplify the expression inside the limit.
We can rewrite \( f(x + h) \):
\[ f(x + h) = \sqrt{3(x + h) + 5} = \sqrt{3x + 3h + 5} \]
Thus, the expression becomes:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{3x + 3h + 5} - \sqrt{3x + 5}}{h} \]

### Step 4: Rationalize the numerator.
To eliminate the square roots, we multiply the numerator and denominator by the conjugate:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{3x + 3h + 5} - \sqrt{3x + 5}\right) \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)}{h \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)} \]

### Step 5: Simplify the numerator.
Using the difference of squares:
\[ (\sqrt{3x + 3h + 5})^2 - (\sqrt{3x + 5})^2 = (3x + 3h + 5) - (3x + 5) = 3h \]
So, the expression simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{3h}{h \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)} \]

### Step 6: Cancel out \( h \) in the numerator and denominator.
\[ f'(x) = \lim_{h \to 0} \frac{3}{\sqrt{3x + 3h + 5} + \sqrt{3x + 5}} \]

### Step 7: Evaluate the limit as \( h \to 0 \).
As \( h \) approaches 0, \( \sqrt{3x + 3h + 5} \) approaches \( \sqrt{3x + 5} \):
\[ f'(x) = \frac{3}{\sqrt{3x + 5} + \sqrt{3x + 5}} = \frac{3}{2\sqrt{3x + 5}} \]

### Final Answer:
Thus, the derivative of \( f(x) = \sqrt{3x + 5} \) is:
\[ f'(x) = \frac{3}{2\sqrt{3x + 5}} \]