camdenreslink
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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}
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}
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