Differentiate by first principle `f(x)=sqrt(3x+4)`

Answer

To differentiate the function \( f(x) = \sqrt{3x + 4} \) using the first principle of derivatives, we will follow these steps:

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

### Step 2: Substitute \( f(x) \) and \( f(x + h) \) into the formula.
We have:
\[ f(x) = \sqrt{3x + 4} \]
Now, calculate \( f(x + h) \):
\[ f(x + h) = \sqrt{3(x + h) + 4} = \sqrt{3x + 3h + 4} \]
Now substitute into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{3x + 3h + 4} - \sqrt{3x + 4}}{h} \]

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

### Step 4: Simplify the numerator.
The numerator simplifies to:
\[ 3h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{3h}{h \left(\sqrt{3x + 3h + 4} + \sqrt{3x + 4}\right)} \]
Cancelling \( h \) in the numerator and denominator gives:
\[ = \lim_{h \to 0} \frac{3}{\sqrt{3x + 3h + 4} + \sqrt{3x + 4}} \]

### Step 5: Apply the limit as \( h \) approaches 0.
As \( h \) approaches 0, \( \sqrt{3x + 3h + 4} \) approaches \( \sqrt{3x + 4} \):
\[ f'(x) = \frac{3}{\sqrt{3x + 4} + \sqrt{3x + 4}} = \frac{3}{2\sqrt{3x + 4}} \]

### Final Answer:
Thus, the derivative of the function \( f(x) = \sqrt{3x + 4} \) is:
\[ f'(x) = \frac{3}{2\sqrt{3x + 4}} \]