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kuruman
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Preface
My first experience with derivatives was seeing how they are obtained from the usual definition
$$f'(x)=\underset{\text{$\Delta $x}\to 0}{\text{Lim}}\frac{f (\text{$x+\Delta $x})-f (x)}{\text{$\Delta $x}}.$$
I accepted the binomial theorem derivation in the case of polynomials and the small angle explanation in the case of sines and cosines  until my math instructor asserted, without justification, that the derivative of the exponential is itself. I had to wait until Taylor expansions in order to understand the exponential derivative. At that time I also saw the series for the cosine and sine, which up to that point I knew as the ratio of one or the other right side to the hypotenuse in a right triangle. I reached graduate school having stored the trigonometric functions, the unit circle, the right triangle side ratios and the series expansions in appropriate compartments in my brain, ready to use them correctly but not really bothering to understand how and if they were...
Continue reading...
 

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  • #2
mathman
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Do you have a question?
 
  • #3
HallsofIvy
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There is no question. This is just a link to his web page.

By the way, you don't have to use Taylor's series to get the derivative of the exponential. With [itex]f(x)= a^x[/itex], [itex]f(x+h)= a^{x+h}= a^xa^h[/itex] so the "difference quotient" is [itex]/frac{a^xa^h- a^x}{h}= a^x\left(\frac{a^h- 1}{h}\right)[/itex]. The derivative is given by [itex]\frac{da^x}{dx}= a^x\left(\lim_{h\to 0} \frac{a^{h}- 1}{h}[/itex]. The limit depends only on a, not on x- it is a constant with respect to x. Writing that constant as [itex]C_a[/itex], the derivative is [itex]C_a a^x[/itex].

Of course, [itex]C_a[/itex] depends on what a is. If, for example, a= 2, the difference quotient is [itex]\frac{2^h- 1}{h}[/itex]. Taking h= 0.001, we get the approximation [itex]\frac{2^{0.001}- 1}{0.001}= \frac{1.0006934- 1}{0.001}= 0.6934[/itex] for [itex]C_2[/itex]. Similarly for a= 3, and taking h= 0.001, [itex]\frac{3^{0.001}- 1}{0.001}= \frac{1.001099- 1}{0.001}= 1.0099[/tex] as an approximation for [itex]C_3[/itex].

Notice that [itex]C_2[/itex] is less than 1 while [itex]C_3[/itex] is larger than 1. So there exist a value of a, between 0 and 1 such that [itex]C_a= 1[/tex]. We define "e" to be equal to that value of a. That is, [tex]frac{d e^x}{dx}= e^x[/itex].
 
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  • #4
HallsofIvy
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I don't know why I was not allowed to edit the previous post. Hopefully, this will come out correctly!

By the way, you don't have to use Taylor's series to get the derivative of the exponential. With [itex]f(x)= a^x[/itex], [itex]f(x+h)= a^{x+h}= a^xa^h[/itex] so the "difference quotient" is [itex]\frac{a^xa^h- a^x}{h}= a^x\left(\frac{a^h- 1}{h}\right)[/itex]. The derivative is given by [itex]\frac{da^x}{dx}= a^x\left(\lim_{h\to 0} \frac{a^{h}- 1}{h}\right)[/itex]. The limit depends only on a, not on x- it is a constant with respect to x. Writing that constant as [itex]C_a[/itex], the derivative is [itex]C_a a^x[/itex].

Of course, [itex]C_a[/itex] depends on what a is. If, for example, a= 2, the difference quotient is [itex]\frac{2^h- 1}{h}[/itex]. Taking h= 0.001, we get the approximation [itex]\frac{2^{0.001}- 1}{0.001}= \frac{1.0006934- 1}{0.001}= 0.6934[/itex] for [itex]C_2[/itex]. Similarly for a= 3, and taking h= 0.001, [itex]\frac{3^{0.001}- 1}{0.001}= \frac{1.001099- 1}{0.001}= 1.0099[/itex] as an approximation for [itex]C_3[/itex].

Notice that [itex]C_2[/itex] is less than 1 while [itex]C_3[/itex] is larger than 1. So there exist a value of a, between 0 and 1 such that [itex]C_a= 1[/itex]. We define "e" to be equal to that value of a. That is, [itex]\frac{d e^x}{dx}= e^x[/itex].
 
  • #5
HallsofIvy
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Another way to handle the exponential is to first define [itex]log(x)= \int_1^x\frac{1}{t}dt[/itex]. We can immediately derive several properties. First, since 1/t is not defined for t= 0, this define log(x) only for x> 0. And then the derivative, 1/x, is positive so log(x) is always increasing. Clearly log(1) is 0 so, since the function is increasing, log(x) is positive for x> 1 and negative for x< 1. It goes to positive infinity for x going to positive infinity and goes to negative infinity for x going to 0.

Most importantly, for our purposes, [itex]log(x^y)= \int_1^{x^y}\frac{1}{t}dt[/itex]. Let [itex]u= t^{1/y}[/itex] so that [itex]t= u^y[/itex] and [itex]dt= yu^{y-1}du[/itex]. When t= 1, u= 1 and when [itex]t= x^y[/itex] [itex]u= x[/tex]. So the integral becomes [itex]\int_1^x\frac{1}{x^y}\left(yu^{y-1}\right)dx= y\int_1^y \frac{1}{u} du= y log(x)[/itex].

Since log(x) is an increasing function, from the positive real numbers to all real numbers, it has an inverse function from all real numbers to the positive real numbers. Call that inverse function "exp(x)". If y= log(x) then x= exp(y) and if y= exp(x) then x= log(y). Since the derivative of log(x) is 1/x, the derivative of y= Exp(x) is 1/(1/y)= y= Exp(x). If y= Exp(x) then x= log(y). If x is not 0, [itex]1= \frac{1}{x}log(y)= log(y^{1/x})[/itex]. Going back the other way, [itex]y^{1/x}= Exp(1)[/tex] so [tex]y= (Exp(1)^x[/itex]. That is, the function, Exp(x), is this number, [itex]Exp(1)[/itex] to the x power. We define "e" to be Exp(1), the number, x, such that log(x)= 1.
 
  • #6
kuruman
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These may be mathematically equivalent ways. However, put yourself in the shoes of a department curriculum committee that has to decide what is taught when and make recommendations that make pedagogical sense. Would you choose one of your two suggested methods or would you recommend waiting until after students have seen series expansions?
 
  • #7
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I liked the hyperreals approach allowing you to skip the notion of limits initially in Calculus. Keisler's Calculus book goes that route and its arguably more intuitive to students initially than limits are.

https://www.math.wisc.edu/~keisler/calc.html
 
  • #8
WWGD
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I liked the hyperreals approach allowing you to skip the notion of limits initially in Calculus. Keisler's Calculus book goes that route and its arguably more intuitive to students initially than limits are.

https://www.math.wisc.edu/~keisler/calc.html
Agreed. Learning is a drill-down process where you first get an overview and gradually break it down until you fully get it. It makes no sense to me to start with a very detailed account.
 
  • #9
TeethWhitener
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Notice that C2C2C_2 is less than 1 while C3C3C_3 is larger than 1. So there exist a value of a, between 0 and 1 such that Ca=1Ca=1C_a= 1. We define "e" to be equal to that value of a. That is, dexdx=exdexdx=ex\frac{d e^x}{dx}= e^x.
If we substitute ##n=1/h## and look for the ##a## for which
$$\lim_{n\to\infty} \left(\frac{a^{1/n}-1}{1/n}\right) = 1$$
with a little rearrangement, we get:
$$a=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n$$
which is another definition of ##e## that many people are familiar with.
 
  • #10
atyy
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There's a similar presentation in John Roe’s “Elementary Geometry” sections 4.1 – 4.4.
https://books.google.com.sg/books?id=Xmb44wyy7gEC
4.1: Dot product
4.2: Pythagorean theorem
4.3: Series definitions of sine and cosine, then defining angle through cosine and the dot product
4.4: SOHCAHTOA
 

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