Relating two properties of $e^x$

Question

1. It is widely known that the derivative of $e^x$ with respect to $x$ is just $e^x$.
2. 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. =)

Answer 1

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.

Answer 2

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

Answer 3

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 $0$.