# Derivative: $e^x$. [duplicate]

How do you differentiate $e^x$?

I looked on many sites, including similar questions here but most answers seemed circular.

The only known definition of $e$ to be used in this proof is $$e=\lim_{n \to\infty} \left(1+\frac{1}n \right)^n$$

What I did is:

\begin{align*} (e^x)' &=\lim_{h\to0}\frac{e^{x+h}-e^x}h \\ &= e^x\lim_{h\to0}\frac{e^{h}-1}h \end{align*}

But I don't know how to go on, I know $\lim_{h\to0}\frac{e^{h}-1}h=1$ but I don't know how to prove it, I can't use the $e^x$ taylor expansion as that would imply diferentiating $e^x$.

Edit: I also can't use the derivative of $\ln(x)$.

• Are you allowed to use Newton's binomial theorem? If so then you just have to expand in binomial series $(1 + 1/n)^{hn}$, substract 1 and find that the only term surviving the limiting process is $h/h =1$, from whence the limit follows. – Rogelio Molina Apr 4 '15 at 0:25
• @RogelioMolina Yes, I'm allowed to use the binomial theorem, could you elaborate on that? I couldn't follow you. – YoTengoUnLCD Apr 4 '15 at 0:26
• See characterizations of the exponential function, especially the paragraph about their equivalence. – Lucian Apr 4 '15 at 0:29
• Another point of interest: You've defined $e$ as a limit. How is $e^h$ defined for all $h\in \mathbb {R}$? – zhw. Apr 4 '15 at 1:42

We must calculate the limit

$$\lim_{h \to 0} \frac{e^h -1}{h}$$ using the definition we have for $e$ we have

$$\lim_{h \to 0}\lim_{n \to \infty}\frac{(1+1/n)^{hn} -1}{h}$$ we expand using the binomial theorem:

$$(1+ 1/n)^{hn} = \sum_{k=0}^{\infty}\binom{hn}{k}(1/n)^k = 1 + h + h^2 \cdots$$ Then the tricky part (this needs to be justified carefully), we exchange the limits to get:

$$\lim_{n \to \infty} \lim_{h\to 0}\frac{1 + h + h^2 \cdots -1}{h} = \lim_{n \to \infty} \lim_{h\to 0}(1 + h\cdots) = \lim_{n \to \infty}1 =1$$

• I thought the binomial theorem was only applicable if we are raising to an integer power. As $h$ grows small, $hn$ may no longer be an integer, right? – layman Apr 4 '15 at 1:03
• The binomial theorem (or a version of it) is valid for all complex powers. See en.wikipedia.org/wiki/Binomial_series – Moya Apr 4 '15 at 1:40
• Even if you don't accept the general binomial theorem, for $h$ rational, there will be a subsequence of $(1+1/n)^{hn}$ where the exponent is an integer, so provided you accept that the original sequence at least converges, there is no loss of generality in assuming $hn$ is integral. – Strants Apr 4 '15 at 1:46
• I'm having trouble seeing why $$\sum_{k=0}^{\infty}\binom{hn}{k}(1/n)^k = 1 + h + h^2 \cdots$$ for the second term I obtain $\frac{h^{2}}{2}-\frac{h}{2n}$ – user135520 Feb 7 '16 at 16:36

Bernoulli's inequality gives that for any $n\geq 2$ we have: $$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n-1}\right)^n\tag{1}$$ hence $n\left(e^{\frac{1}{n}}-1\right)$ is between: $$1 < n\left(e^{\frac{1}{n}}-1\right) < 1+\frac{1}{n-1}\tag{2}$$ hence, by squeezing: $$\lim_{n\to +\infty}\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}=1 \tag{3}$$ that, together with $(2)$, gives: $$\lim_{r \to 0}\frac{e^r-1}{r}=1\tag{4}$$ that is enough to grant: $$\lim_{h\to 0}\frac{e^{x+h}-e^{x}}{h}=e^x \tag{5}$$ as wanted.

• (+1) I was going to go this way before I saw your answer. I am just too slow in my old age! Bernoulli's Inequality is so very useful. – robjohn Apr 4 '15 at 1:57
• I hope you don't mind. I've added an answer to add a bit more detail about how $(1)$ is derived from Bernoulli's Inequality. – robjohn Apr 4 '15 at 2:14

This is really a comment on Jack D'Aurizio's answer, but it is too long for a comment. Let me show how Bernoulli's Inequality is used to prove $(1)$ from Jack D'Aurizio's answer.

In this answer, it is shown, using Bernoulli's Inequality, that $$\left(1+\frac1n\right)^n\tag{1}$$ is an increasing sequence and that $$\left(1+\frac1n\right)^{n+1}\tag{2}$$ is a decreasing sequence. This means that for all $n$ $$\left(1+\frac1n\right)^n\le\overbrace{\lim_{k\to\infty}\left(1+\frac1k\right)^k}^{\large e}\le\left(1+\frac1n\right)^{n+1}\tag{3}$$ Since $(3)$ is true for all $n$, we can substitute $n\mapsto n-1$ in the right-side inequality to get the inequality from Jack's answer: $$\left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{4}$$

\begin{equation*} e^x=\lim_{n\rightarrow \infty}\left(1+\frac {x} {n}\right)^n =\lim_{n\rightarrow \infty}\sum_{k=0}^n\binom n k\frac {x^k} {n^k} =\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n(n-1)...(n-k+1)} {k!}\frac {x^k} {n^k}\end{equation*} $$=\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n(n-1)...(n-k+1)} {n^k}\frac {x^k} {k!}=\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n} {n}\frac {n-1}{n}...\frac{n-k+1} {n}\frac {x^k} {k!}$$ $$\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {1} {1}\left(1-\frac {1}{n}\right)...\left(1-\frac{k-1} {n}\right)\frac {x^k} {k!}=\sum_{k=0}^\infty\frac {x^k} {k!}$$ Hence we have that $$e^x=\sum_{k=0}^\infty\frac {x^k} {k!}\implies(e^x)'=\sum_{k=1}^\infty\frac{kx^{k-1}} {k!}=\sum_{k=1}^\infty\frac{x^{k-1}} {(k-1)!}=\sum_{k=0}^\infty\frac{x^{k}} {k!}$$ It follows that $e^x=(e^x)'$

Define the log function as $\log x=\int_1^x \frac{dt}{t}$. The derivative is obviously $1/x$. The exponential function is the inverse of the log function. So, we may write

\begin{align} x&=\log (e^x)\\ &=\int_1^{e^x} \frac{dt}{t} \end{align}

Taking the derivative of both sides and applying the chain rule reveals

\begin{align} \frac{dx}{dx}&=1\\ &=\frac{d\log(e^x)}{dx}\\ &=\frac{1}{e^x}\frac{de^x}{dx} \end{align}

whereupon solving for $\frac{de^x}{dx}$ shows that

$$\frac{de^x}{dx}=e^x$$

So, the derivative of the exponential function is itself!

$$e^x=\sum_{n\ge 0} \frac{x^n}{n!}$$ $$\frac{de^x}{dx}=\frac{d}{dx}[1+x+x^2/2!+x^3/3!+\cdots]=0+1+x+x^2/2+\cdots=\sum_{n\ge 0} \frac{x^n}{n!}=e^x$$

This uses a theorem justifying interchanging the derivative with the summation.

• From the original post: but I don't know how to prove it, I can't use the ex Taylor expansion as that would imply differentiating $e^x$. – MathMajor Apr 4 '15 at 1:34
• @GabrielH The Taylor expansion only requires knowing the derivative at a single point, and that can be easily found by approximation with secant lines. – Teoc Apr 4 '15 at 1:35
• The OP has likely not even learned that term by term differentiation of an infinite series is allowed ... – MathMajor Apr 4 '15 at 1:36
• It is the OP's problem for not including enough context. – Teoc Apr 4 '15 at 1:37
• Okay, that may be the case. It wasn't me who down voted. I just pointed out that answer is most likely not helpful to the OP. – MathMajor Apr 4 '15 at 1:37