To solve these problems using the increment method of differentiation, also known as the difference quotient method, we use the definition of the derivative:

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

I'll walk you through the steps to solve each problem, starting with the first. Let's tackle these one by one.

1. $f(x) = 3 - 2x + 4x^2$

Using the difference quotient method: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

• First, find $f(x+h)$: $f(x+h) = 3 - 2(x+h) + 4(x+h)^2 = 3 - 2x - 2h + 4(x^2 + 2xh + h^2)$ $= 3 - 2x - 2h + 4x^2 + 8xh + 4h^2$

Now, compute the difference quotient: $f'(x) = \lim_{h \to 0} \frac{(3 - 2x - 2h + 4x^2 + 8xh + 4h^2) - (3 - 2x + 4x^2)}{h}$ $= \lim_{h \to 0} \frac{-2h + 8xh + 4h^2}{h}$ Simplify: $f'(x) = \lim_{h \to 0} (-2 + 8x + 4h) = -2 + 8x$ Thus, $f'(x) = -2 + 8x$.

If you want, I can solve each derivative similarly, step by step. Would you like me to continue with the others, or would you prefer to solve them together?

Also, here are five questions to consider for expanding your understanding:

1. How does the increment method differ from other differentiation methods?
2. What is the geometric interpretation of the derivative?
3. Can you apply the difference quotient to trigonometric functions?
4. What happens if the limit in the difference quotient doesn't exist?
5. Why does the limit $h \to 0$ lead to the instantaneous rate of change?

Tip: Practice applying the difference quotient to different functions to strengthen your understanding of derivatives.