I) Define derivatives of a function from first principle. ii) Find the derivatives of f(x) = 3x2 − 5x from first principle.

The derivative of a function $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{dy}{dx}$, is defined as the limit of the average rate of change of the function as the change in $x$ approaches zero. It represents the instantaneous rate of change of the function at a given point.
The formula for the derivative from first principles is:

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

Given $f(x) = 3x^2 - 5x$.

First, find $f(x+h)$:

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

Expand $(x+h)^2$:

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

Next, find $f(x+h) - f(x)$:

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

Cancel out terms:

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

Now, divide by $h$:

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

Factor out $h$ from the numerator:

$$\frac{h(6x + 3h - 5)}{h}$$

Cancel $h$ (since $h \neq 0$ as $h \to 0$):

$$6x + 3h - 5$$

Finally, take the limit as $h \to 0$:

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

Substitute $h=0$:

$$f'(x) = 6x + 3(0) - 5$$ $$f'(x) = 6x - 5$$

The derivative of $f(x) = 3x^2 - 5x$ from first principle is $\boxed{6x - 5}$.