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: Define the function and the limit expression
According to the first principle of derivatives, the derivative \( f'(x) \) can be expressed as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Here, \( f(x) = \sqrt{3x + 5} \).

### Step 2: Substitute \( f(x+h) \) and \( f(x) \)
Now, we need to compute \( f(x+h) \):
\[ f(x+h) = \sqrt{3(x+h) + 5} = \sqrt{3x + 3h + 5} \]
Thus, we can write:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{3x + 3h + 5} - \sqrt{3x + 5}}{h} \]

### Step 3: Multiply by the conjugate
To simplify the expression, we multiply the numerator and denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{3x + 3h + 5} - \sqrt{3x + 5}\right) \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)}{h \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)} \]
This gives us:
\[ = \lim_{h \to 0} \frac{(3x + 3h + 5) - (3x + 5)}{h \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)} \]

### Step 4: Simplify the numerator
The numerator simplifies to:
\[ 3h \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{3h}{h \left(\sqrt{3x + 3h + 5} + \sqrt{3x + 5}\right)} \]

### Step 5: Cancel \( h \)
We can cancel \( h \) from the numerator and denominator:
\[ f'(x) = \lim_{h \to 0} \frac{3}{\sqrt{3x + 3h + 5} + \sqrt{3x + 5}} \]

### Step 6: Evaluate the limit
Now, we can substitute \( h = 0 \):
\[ f'(x) = \frac{3}{\sqrt{3x + 5} + \sqrt{3x + 5}} = \frac{3}{2\sqrt{3x + 5}} \]

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

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