Derivative of $e^{x}$ by definition

Question

if you plug in $e^{x}$ into the definition of a derivative, you'll end up with

the limit $\lim_{h\to 0}\frac{e^{h-1}}{h}$. This limit is a ''standard limit'' and equals 1.

However if i wouldnt know this is a standard limit that equals 1, but use a taylor series instead to evaluate that limit, would that be justified? because the coefficient of a taylor series of a function involves a derivative, you're making use of something that you actually dont know yet (namely the derivative of $e^{x}$)

So the question is: is a taylor series justified to use in the definition of a derivative?

@OdellBeckhamJr Taylor's formula requires the computation of derivatives of the exponential... and you want to deduce the derivative of the exponential... — PierreCarre, May 05 '20 at 13:38
The usual definition of the exponential function is done as a power series or as the inverse function of the integral over $1/x$. Which one do you prefer? — Michael Hoppe, May 05 '20 at 13:43
Does this answer to your question: https://math.stackexchange.com/questions/359023/using-the-limit-definition-to-find-the-derivative-of-ex — VIVID, May 05 '20 at 13:54
@PierreCarre yeah, my question is: is it justified to use a taylor series? — iSauceNervouzWithoutHezzy, May 05 '20 at 14:02
You can define the exponential function via $\exp(x) =\sum_{n=0}^{\infty} x^n/n! $ and then prove that $\exp(x+y) =\exp(x) \exp(y) $ and $\exp'(x) =\exp(x) $. Using these one proves that it can work as an exponential $e^x$ where $e=\exp(1)$ and further that the original defining series indeed equals its Taylor series. There is no circularity involved here. And this is one of easier approaches to theory of exponential function. — Paramanand Singh, May 05 '20 at 14:09
By the way any definition you choose for the exponential function should allow you to prove all the properties without any circularity. And it is better to know multiple approaches. — Paramanand Singh, May 05 '20 at 14:12
I guess the question is not specifically about $e^x$ but in general about $a^x$. — Gyro Gearloose, May 05 '20 at 14:12
This might help : https://math.stackexchange.com/questions/1491220/show-lim-h-to-0-fracah-1h-exists-without-lh%c3%b4pital-or-even-referenc — Arnaud D., May 07 '20 at 09:16

score 1 · Answer 1 · answered May 05 '20 at 13:54

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If you define the exponential as $\displaystyle e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$, and differentiate term by term (you can prove this is ok...) you get $$ (e^x)' = \left(1+x+x^2/2 + x^3/3! + \cdots\right)' = 1+x+x^2/2 +x^3/3! + \cdots) = e^x.$$

answered May 05 '20 at 13:54

PierreCarre

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score 1 · Answer 2 · answered May 05 '20 at 14:01

Here's how I approach these topics. (i) Define a function $\exp:\mathbb R \to \mathbb R$ by $$\exp(x)=1+\sum_{i=1}^{\infty}\frac{x^n}{n!}.$$ (ii) Differentiate this defining equation to prove that $$\exp'(x)=\exp(x)$$(iii) Define $$e=\exp(1).$$ (iv) Prove $$\exp(x)=e^x \quad \forall x \in \mathbb R.$$ (v) Prove that $$e=\lim_{n \to \infty}(1+\frac{1}{n})^n.$$

JayP · Answer 3 · 2020-05-05T13:51:19.073

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The definition of the limit says; $$f^\prime (x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ Given that we care considering $f(x)=e^x$ then $f(x+h)=e^{x+h}$.Substituting this into the expression; $$f^\prime (x)=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$$ This can be simplified; $$f^\prime (x)= e^{x}\lim_{h\to 0}\frac{(e^h-1)}{h}$$ Refer to; How to prove that $\lim_{h\to 0} \, \frac{e^h-1}{h}$ Then; $$f^\prime (x)= e^{x}\lim_{h\to 0}\frac{(e^h-1)}{h}=e^x \times 1$$ And thus; $$f^\prime(x)=e^x$$

edited May 05 '20 at 13:51

answered May 05 '20 at 13:42

JayP

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Doesn't l'Hopital use the derivative that you are trying to proof? – Gyro Gearloose May 05 '20 at 13:43
@MATHBOI You took a long walk to get to the same spot! – PierreCarre May 05 '20 at 13:46
Modified my response @GyroGearloose Thank you! – JayP May 05 '20 at 13:56
i know that that limit equals 1. but my question is: if you use a taylor series to evaluate that limit, would that be valid, since a taylor series makes use of a derivative itself. you're basically using something that you dont know yet – iSauceNervouzWithoutHezzy May 05 '20 at 14:00

Derivative of $e^{x}$ by definition

3 Answers3