We are asked to differentiate the function $y = 3x^2 + 2x + 5$ from the first principle.

1. Problem Description

We are asked to differentiate the function $y = 3x^2 + 2x + 5$ from the first principle.

2. Solution Steps

The first principle of differentiation states that:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Here, $f(x) = 3x^2 + 2x + 5$. Thus, we need to find $f(x+h)$:

$$f(x+h) = 3(x+h)^2 + 2(x+h) + 5$$

$$f(x+h) = 3(x^2 + 2xh + h^2) + 2x + 2h + 5$$

$$f(x+h) = 3x^2 + 6xh + 3h^2 + 2x + 2h + 5$$

Now we substitute $f(x+h)$ and $f(x)$ into the first principle formula:

$$f'(x) = \lim_{h \to 0} \frac{(3x^2 + 6xh + 3h^2 + 2x + 2h + 5) - (3x^2 + 2x + 5)}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 + 2x + 2h + 5 - 3x^2 - 2x - 5}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2 + 2h}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{h(6x + 3h + 2)}{h}$$

$$f'(x) = \lim_{h \to 0} (6x + 3h + 2)$$

$$f'(x) = 6x + 3(0) + 2$$

$$f'(x) = 6x + 2$$

3. Final Answer

The derivative of $y = 3x^2 + 2x + 5$ from the first principle is $6x + 2$.