Derivative: $e^x$.

Question

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 $\ln(x)$.

Answer 1

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:

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

Answer 2

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.

Answer 3

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: $$ \left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{4} $$

Answer 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 $e^x=(e^x)'$

Answer 5

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!

Answer 6

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

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