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This question already has an answer here:

Using the definition of derivative, one can find that $\frac{d}{dx} (e^x) = e^x$ using the fact that $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$. Fine.

However, the proofs that I have seen of that latter fact have all used the power series $e^x = 1 + x + x^2/2 + \dots$ which, to my knowledge, require all derivatives of $e^x$ to compute. So, is this proof circular? And if so, how can we correctly show that $\frac{d}{dx} (e^x) = e^x$ using the definition of a derivative?

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marked as duplicate by Ali Caglayan, Aaron Maroja, TMM, Chappers, Mark Fantini Apr 8 '15 at 0:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You may find this useful: math.stackexchange.com/questions/359023/… $\endgroup$ – Radz May 31 '14 at 16:53
  • $\begingroup$ This comment is not directed at OP but rather to everyone. If you take $e^x=\sum \frac{x^k}{k!}$ as your definition how will you prove that $e^x$ is equal to the number $e$ raised to the power $x$ ? $\endgroup$ – Rene Schipperus May 31 '14 at 17:15
  • $\begingroup$ Also, you might look at math.stackexchange.com/a/199470/36530. I actually define the number $e$ such that $\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1$ to implicitly define $e$. If you used the power series to define the exponential function, then you could define $e$ as $f(1)$. There are many options really. $\endgroup$ – James S. Cook May 31 '14 at 18:42
  • $\begingroup$ Probably what is confusing about the series is that you know you are putting the values of the derivatives at a given point into all the coefficients, so you think you need the derivatives. That is not true. Starting with the series is like saying "Define $e^x$ as the function where all derivatives at 0 are 1." Now you are asking "What is the derivative of this function at all points?" You weren't given that, so it's not circular. If you find it seems too easy, that is good. It means you are comfortable with series, and they should be easy and fun. $\endgroup$ – ex0du5 Jun 1 '14 at 0:05
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There are many, many different ways you can approach this.

(1) Define $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$. Then taking a derivative and passing the limit through (this needs to be justified but it can be done by showing uniform convergence) we can show that $\frac{d}{dx}e^x = e^x$. Alternatively, we can use the limit definition of a derivative and this definition to show this.

(2) Define $e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$. See Rene's answer for finding the bound $$\limsup_{h \to 0} \frac{e^h - 1}{h} \leq \frac{k+1}{k}$$ to see how we can compute the limit and hence the derivative.

(3) Define $e^x$ to be the number $y(x)$ such that $\int_1^{y(x)} \frac{1}{t} dt = x$. Use the fundamental theorem of calculus and the chain rule when taking a derivative of the above with respective to $x$ to get $$\frac{1}{y(x)} y'(x) = 1$$ and conclude $y'(x) = y(x)$ or $(e^x)' = e^x$.


Now all these definitions are equivalent. That is, starting with one definition, we can show that the other must hold. It is clear that (2) and (3) imply (1) because knowing the derivative of $e^x$ you can find it's taylor series. But this then means that (2) and (3) give the same function. So they are all equivalent.

The truly circular way to compute the limit of the finite difference is to use L'Hospital's rule: $$\lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{e^h}{1} = 1.$$ The problem is that you use what the derivative of $e^h$ is to show what the derivative of $e^x$ is.

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    $\begingroup$ I used to think I was smart because I calculated $\lim_{x \rightarrow 0} \frac{\sin x}{x}=1$ by L'Hopital's rule and these other people were doing all this complicated geometry stuff... then, I proved the derivative of $\sin x$ was $\cos x$ from the definition... $\endgroup$ – James S. Cook May 31 '14 at 18:44
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Well I dont know if this is the answer you want, but here is a proof of $$\lim\limits_{x \to 0} \frac{e^{x}-1}{x}=1$$ direct from the definition of $e^x$ as limit.

Note that $\left(1+\frac{x}{k} \right)^{k+1}$ is a decreasing sequence (induction, and Bernoulli inequality) and its limit as $k\to \infty$ is $e^x$ by the definition of $e^x$. Now we have

$$1\leq \frac{e^{x}-1}{x} \leq \frac{1}{x}\left[\left( 1+\frac{x}{k} \right)^{k+1} -1\right]$$ for all $k$. And the latter has limit $\frac{k+1}{k}$ as $x\to 0$.

So $$1 \leq \lim\limits_{x \to 0} \frac{e^{x}-1}{x} \leq \frac{k+1}{k}$$ for all $k$.

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  • $\begingroup$ I'm missing why it is that $\frac{1}{x}\left[\left(1+\frac{x}{k}\right)^{k+1}-1\right]$ has limit $\frac{k+1}{k}$ as $k\rightarrow \infty$. Can help with that? [EDIT: Sorry, just noticed $x\rightarrow \infty$, nevermind, got it.] $\endgroup$ – Addem May 31 '14 at 17:11
  • $\begingroup$ OK I give my proof anyway, $\frac{1}{x}\left[\left( 1+\frac{x}{k} \right)^{k+1} -1\right]=\frac{1}{k}((1+\frac{x}{k})^k+(1+\frac{x}{k})^{k-1}+\cdots +1) $ now $x\to 0$. $\endgroup$ – Rene Schipperus May 31 '14 at 17:19
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No: the definition $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!} $$ is independent of anything but convergence of series. You first prove that is converges for any real $x$, then you prove that the derivatives coincide with $e^x$ itself.

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To find the derivative using the power series notation... $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$ \begin{align} \frac{d}{dx} e^x &= \sum_{k=1}^{\infty} \frac{kx^{k-1}}{k!} \\ &= \sum_{k=0}^{\infty} \frac{(k+1)x^{k}}{(k+1)!} & \text{Transform }k-1 \mapsto k \\ &= \sum_{k=0}^{\infty} \frac{(k+1)x^{k}}{(k+1)k!} \\ &= \sum_{k=0}^{\infty} \frac{x^{k}}{k!} = e^x \end{align}

Personally, I do not see how this is circular reasoning for the definition of the derivative. The Maclaurin series is just another representation of $e^x$, and this allows calculating $\frac{d}{dx} e^x$ by using the power rule.

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