Differentiate by `1^(st)` principle : `f(x)=sqrt(3x-4)`

Answer

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

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

### Step 2: Substitute \( f(x) \) and \( f(x + h) \).
We have:
\[ f(x) = \sqrt{3x - 4} \]
Now, calculate \( f(x + h) \):
\[ f(x + h) = \sqrt{3(x + h) - 4} = \sqrt{3x + 3h - 4} \]
Now, substitute these 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 will multiply and divide by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{3x + 3h - 4} - \sqrt{3x - 4}\right) \left(\sqrt{3x + 3h - 4} + \sqrt{3x - 4}\right)}{h \left(\sqrt{3x + 3h - 4} + \sqrt{3x - 4}\right)} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{(3x + 3h - 4) - (3x - 4)}{h \left(\sqrt{3x + 3h - 4} + \sqrt{3x - 4}\right)} \]

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

### Step 5: Cancel \( h \) in the numerator and denominator.
Assuming \( h \neq 0 \), we can cancel \( h \):
\[ f'(x) = \lim_{h \to 0} \frac{3}{\sqrt{3x + 3h - 4} + \sqrt{3x - 4}} \]

### Step 6: Evaluate 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:
\[ f'(x) = \frac{3}{2\sqrt{3x - 4}} \]