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

You're on a roll — let's tackle question 4.d. Question 4.d has two parts: i) Define derivatives of a function from first principle. ii) Find the derivatives of f(x) = 3x^2 - 5x from first principle. Step 1: Define derivatives of a function from first principle. The derivative of a function f(x) with respect to x, denoted as f'(x) or (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) = _h 0 (f(x+h) - f(x))/(h) Step 2: Find the derivatives of f(x) = 3x^2 - 5x from first principle. 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: (f(x+h) - f(x))/(h) = (6xh + 3h^2 - 5h)/(h) Factor out h from the numerator: (h(6x + 3h - 5))/(h) Cancel h (since h ≠ 0 as h 0): 6x + 3h - 5 Finally, take the limit as h 0: f'(x) = _h 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 6x - 5. What's next? Send 'em.