Relating two properties of $e^x$

Question

It is widely known that the derivative of $e^x$ with respect to $x$ is just $e^x$.
It is commonly known that the slope of the tangent line of $e^x$ at the point $x=0$ is $1$.

How can I relate 1. with 2. ? That is how can I explain 1. using 2. ? My intuition says that they are related but I can't figure out how!!

In other words: how can I show that $\frac{d}{dx} e^x = e^x$ from the fact that the slope of the tangent line of $e^x$ at the point $x=0$ is $1$. (my feeling tends to see that $1 \times e^x$ = $e^x$ may explain it)

'$e^x$ (where $x$ is the number you think about now)' thanks. =)

Do you want: as $h\rightarrow0$ $${e^{x+h}-e^x\over h}=e^x\cdot{e^h-1\over h}\longrightarrow e^x\cdot1=e^x?$$ — David Mitra, Dec 18 '13 at 20:28
Are you asking to find a function, $f$ such that $f'(x) = f(x)$? Your question is confusing. — muffle, Dec 18 '13 at 20:37
no Mr. muffle my question is: how can I derive that d/dx e^x = e^x from the fact that the slope of the tangent line of e^x at the point x=0 is 1. — user123456789, Dec 18 '13 at 20:44

score 1 · Answer 1 · answered Dec 18 '13 at 20:24

1

$e^0=1$, and this is the slope of $e^x$ at $x=0$. So...

answered Dec 18 '13 at 20:24

Ian Mateus

7,431

John · Accepted Answer · 2013-12-18T20:55:51.263

0

If $f(x) = e^x$, then $f(0) = e^0 = 1$.

And $f'(0) = e^0 =1$.

And $f''(0) = e^0 =1$.

And $f'''(0) = e^0 =1$.

And ...

But this is just a property of the exponential function.

You can draw no conclusions about $f(x)$ simply from one value of the first derivative. The fact that $f'(0) = 1$ "looks nice" is just a happy coincidence.

Consider

$$g(x) = \frac{(x+1)^2}{2}.$$

Now, $g'(x) = x+1$, so $g'(0) = 1.$ But clearly $g'(x) \neq g(x)$. So you cannot say anything about the equality (or inequality) of a function and its derivative simply knowing the value of the derivative at one point. (Or even many points!)

The exponential function is special because its derivative is the function itself for all x. Showing that it's the same at once point isn't sufficient.

edited Dec 18 '13 at 20:55

answered Dec 18 '13 at 20:27

John

26,319

So if slope of function g(x) for example at a point z for instance is 1 and if g(z)=c then g'(z)=c and g''(z)=c .... True? – user123456789 Dec 18 '13 at 20:28
If $g(x) = e^x$, yes. – John Dec 18 '13 at 20:29
But how did you get f'(0)=1 then from 1. ?? I don't understand, if the slope at x=0 is 1 what this means? – user123456789 Dec 18 '13 at 20:30
john you used 1. (that d/dx e^x = e^x)in attempt to explain 1. which is circular argument. I'm searching for d/dx e^x=e^x derivation only from 2. – user123456789 Dec 18 '13 at 20:33
I used the fact that $f'(x) = f(x)$ if $f(x) = e^x.$ Isn't that your first statement in your question? – John Dec 18 '13 at 20:34
you can't use it since i want to derive it from 2. !! – user123456789 Dec 18 '13 at 20:34
Oh, I see. You can't explain the first statement using the second. Consider $f(x) = (x+1)^2/2$. $f'(0) = 1$, but clearly $f'(x) \neq f(x)$. – John Dec 18 '13 at 20:38
But the slope of the tangent line of f(x)=(x+1)^2/2 at point x=0 is 1? sure? – user123456789 Dec 18 '13 at 20:46

Tim Ratigan · Answer 3 · 2013-12-18T21:07:58.770

A proof that $\frac{\text d}{\text dx}e^x=e^x$ might help:

Let $f(x)$ be a function such that $f'(x)=f(x)$. Then $$ f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} $$

For this to be true, it would be really nice for $f(x)$ to satisfy $f(x+h)=f(x)f(h)$ so that we can "factor out" $f(x)$ (there's a bit more reasoning behind this step but let's give intuition credit and move on). Now, there are only two types of functions that satisfy the functional equation $f(a+b)=f(a)f(b)$ and those are $f(x)=0$ and $f(x)=k^x$ for some $k$ (This is illustrated very nicely in Christopher Small's Functional Equations and How to Solve Them, if you're interested.). We don't want the boring old $0$ function, so let's call our function $f(x)=e^x$ for an arbitrary $e$. Then we have:

$$\begin{align}e^x&=\lim_{h\to0}\frac{e^xe^h-e^x}{h}\\ 1&=\lim_{h\to0}\frac{e^h-1}{h}\tag 1\\ \lim_{h\to0}(1+h)^{1/h}&=e\\ e&=\lim_{x\to\infty}\left(1+\frac1x\right)^x\end{align}$$ Now, what does $(1)$ require of $e^x$? That's right! Your property $2$ (or some version of it), that $e^x-1$ is arbitrarily close to $x$ around $0$.

Relating two properties of $e^x$

3 Answers3