We need to differentiate the function $y = 3x^2 + 2x + 5$ from first principles.

1. Problem Description

We need to differentiate the function $y = 3x^2 + 2x + 5$ from first principles.

2. Solution Steps

Differentiation from first principles uses the following formula:

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

Let $f(x) = 3x^2 + 2x + 5$. Then, $f(x+h) = 3(x+h)^2 + 2(x+h) + 5$.

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

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

Now, we calculate $f(x+h) - f(x)$:

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

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

Next, we divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 + 2h}{h} = 6x + 3h + 2$

Finally, we take the limit as $h$ approaches 0:

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

3. Final Answer

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