Derivatives: The Method of Increments

[camdenreslink]

I am using this book: https://www.amazon.com/dp/0486404536/?tag=pfamazon01-20
Calculus: An Intuitive and Physical Approach by: Kline

I've already finished Calc I and II, but I'm brushing up on my skills because I don't feel entirely confident about them.

This is what the method says to do to find a derivative:

$$s = 10t^2$$ given $$t=3$$

$$s_{3} + \Delta s = 10(3 + \Delta t)^2$$

$$s_{3} + \Delta s = 90 + 60\Delta t + 10\Delta t^2$$
$$- (s_{3} = 90)$$


$$\frac{\Delta s}{\Delta t} = \frac{60 \Delta t + 10 \Delta t^2}{\Delta t}$$

$$\displaystyle{\frac{\Delta s}{\Delta t}} = 60 + 10 \Delta t$$

$$lim_{\Delta t \rightarrow 0} \displaystyle{\frac{\Delta s}{\Delta t}} = 60$$

I understand the method and have finished all of the practice problems in the book, but I'm having trouble linking this with the way I was taught to derive in my calculus class with

$$lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$$

 

[Mark44]

I really hate his abuse of notation, using s₃ for s(3). There seem to be a lot of steps missing, which might make it difficult to follow his logic.

For example, $\Delta s$ = s(3 + $\Delta t$) - s(3) = 10(3 + $\Delta t$)² - 90
= 90 + 60 $\Delta t$ + 10 ( $\Delta t$)² - 90
= 60 $\Delta t$ + 10 ( $\Delta t$)²

Dividing by delta t gives you this, the same as above:
$$\frac{\Delta s}{\Delta t}} = 60 + 10 \Delta t$$

Finally, take the limit to get ds/dt.

Delta t in his exposition is the same as h in what you're used to. Hopefully, my explanation will help you understand the parallels between the two approaches.