Show that if $ f'(0)=1, f'(x)=e^x $ with the law of exponents I'm really stuck on this one. Show that, if $f'(0)=1$ is known, one can deduce
$
f'(x)=e^x
$ with the aid of the law of exponents.
I started here and assumed that I should go towards L'hospital's rule but I get stuck after.
$\lim \limits_{x \to 0} \frac{f(x) - f(0)}{x}
$
The wording exactly: 
Let $f(x)=e^x$. Explain why $ f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$. Show that, if $f'(0)=1$ is known, one can deduce $f'(x)=e^x$ with the aid of the laws of exponents.
The first part was trivially easy but the way the problem is worded makes it seem like two separate sets of facts to deduce from. 
 A: EDIT: To clarify for you, the problem is asking you to derive the fact that the derivative of $e^x$ is $e^x$ from the basic laws of exponents, rather than as an inherent rule of derivatives.
Given $f(x) = e^x$ we see that $f'(a) = \lim\limits_{x \to a} \dfrac{e^x - e^a}{x-a}$. Therefore $f'(0) = \lim\limits_{x \to 0} \dfrac{e^x - e^0}{x-0} = \lim\limits_{x \to 0}\dfrac{e^x - 1}{x}$. Given that $f'(0) = 1$, we naturally see $\lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$.
So we now know $f(x) = e^x$ and $f'(0) = \lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$. Use these facts to show $f'(x) = e^x$. This can be done as follows:
The general form of a derivative $f'(x) = \lim\limits_{h \to 0}\dfrac{f(x+h) - f(x)}{h}$. In this case, $f'(x) = \lim\limits_{h \to 0}\dfrac{e^{x+h}-e^x}{h}$. By the properties of exponents, $f'(x) = \lim\limits_{h \to 0}\dfrac{e^xe^h-e^x}{h}=\lim\limits_{h \to 0}\dfrac{e^x(e^h - 1)}{h}$.
Because the limit is with respect to $h$ (not $x$), $e^x$ is a constant, and can be pulled out of the limit such that $f'(x) = e^x \lim\limits_{h \to 0} \dfrac{e^h - 1}{h}$. Substituting $h$ for $x$ in the given fact that $\lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$ we see $f'(x) = e^x (1) = e^x$.
A: Note that $f'(x) = x+1\neq e^x$ satisfies the condition $f'(0) = 1$, but $f(x) = \dfrac{x^2}{2}+x\neq e^x$. This should be thrown in the comment but I accidentally posted here. 
A: 
Let $f(x)=e^x$. Explain why $ f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$. Show that, if $f'(0)=1$ is known, one can deduce $f'(x)=e^x$ with the aid of the laws of exponents.

Here's how I would parse this problem statement:

Let $f(x)=e^x$.

This defines $f$ for all the remaining parts of the problem.

Explain why $f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$.

As you already inferred, this part of the problem refers to the same $f$
that was defined at the start of the problem. You need to prove that this
equation is true. (As you noticed, this is easy.)

Show that, if $f'(0)=1$ is known,

When you get to this part of the problem, you will have shown in the previous part of the problem that $f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$.
But you will not have shown that $\lim \limits_{x \to 0} \frac{e^x - 1}{x} = 1$.
The words "if $f'(0)=1$ is known" allow you to assume this fact
without proving it; it is given to you by the problem statement.

one can deduce $f'(x)=e^x$ with the aid of the laws of exponents.

Now that you know that $f(x)=e^x$ (by definition in the problem statement), that
$f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$ (already proved by you),
and that $\lim \limits_{x \to 0} \frac{e^x - 1}{x} = 1$
(additional information given by the problem statement),
the task is to show that $f'(x) = e^x$.
A: if you assume the $f(x)=a^x$ then you can show that $f'(0)=1\implies a=e$
$$f'(x)= \lim _{h \to 0}\frac{a^{(x+h)}-a^x}{h} = a^x \lim _{h \to 0}\frac{a^h-1}{h} $$
So
$$ f'(0)=1 \implies \lim _{h \to 0}\frac{a^h-1}{h}=1 $$
Now $g(a)= \lim _{h \to 0}\frac{a^h-1}{h}$ is a perfectly well-defined function of $a$ whose inverse function is given by...
$$g^{-1}(x) = \lim _{h \to 0} (1+h x )^{1/h}$$
So 
$$ f'(0)=1 \implies g(a)=1 \implies a = g^{-1}(1) = \lim _{h \to 0} (1+h )^{1/h}=e$$
A: There exist an infinite number of functions whose derivative, at x= 0, is 1. It is certainly NOT true that the derivative must be f'(x)= e^x.
