Proof of derivative of $e^x$ is $e^x$ without using chain rule Is there a way to prove that the derivative of $e^x$ is $e^x$ without using chain rule? If so, what is it? Thanks.
 A: Hmmm.... Well, how precisely have you defined $e^x$? Depending on the answer, the approach will vary. If you've defined $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!},$$ then it will follow fairly readily that $e^x$ is its own derivative, using Taylor series properties.
If on the other hand you've defined $$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n,$$ then you may have a slightly harder way to go. I think using the difference of $n$th powers formula may help.
A: To follow up on "what is your definition of $e^x$", if your definition is as the solution to the differential equation $y'=y$ such that $y(0)=1$, then you have nothing to prove!
A: The discovery of the constant $e$ is credited to Jacob Bernoulli in 1683 who attempted to find the value of the following expression (which is equal to $e$):
$$\lim_{n\to \infty}{\left(1+\frac1n\right)}^n.$$
Alternatively, we can substitute $n=\frac1h$ to obtain:
$$e=\lim_{h\to0}(1+h)^{1/h}.$$
If $y=e^x$ then, from first principles:
$$\frac{dy}{dx}=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$$
$$=e^x\lim_{h\to0}\frac{e^h-1}h,$$
as the value of $e^x$ does not depend on $h$. Now we substitute our limit from earlier to get:
$$\frac{dy}{dx}=e^x\lim_{h\to0}\frac{{\left((1+h)^{1/h}\right)}^h-1}h$$
$$=e^x\lim_{h\to0}\frac{{(1+h)}-1}h$$
$$=e^x.$$
A: When using the definition
$$\mathrm e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
you can proceed as follows:
$$\begin{aligned}
\frac{\mathrm d}{\mathrm dx}\mathrm e^x
&= \lim_{h\to 0}\frac{\mathrm e^{x+h}-\mathrm e^x}{h}\\
&= \lim_{h\to 0}\frac{\lim\limits_{n\to\infty}\left(1+\frac{x+h}n\right)^n - \lim\limits_{n\to\infty}\left(1+\frac xn\right)^n}{h}\\
&= \lim_{h\to 0}\lim_{n\to\infty}\frac{\left(1+\frac{x+h}n\right)^n - \left(1+\frac xn\right)^n}{h}
\end{aligned}$$
 Now
$$\left(1+\frac{x+h}{n}\right)^n = \sum_{k=0}^n{n\choose k}\left(\frac{h}{n}\right)^k\left(1+\frac{x}{n}\right)^{n-k}$$
and therefore
$$\begin{aligned}
\frac{\mathrm d}{\mathrm dx}\mathrm e^x
&= \lim_{h\to 0}\lim_{n\to\infty}\sum_{k=1}^n{n\choose k}\frac{h^{k-1}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\\
&= \lim_{h\to 0}\left(\lim_{n\to\infty}{n\choose 1}\frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1}+h\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\right)\\
&= \lim_{n\to\infty}{n\choose 1}\frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1}+\lim_{h\to 0}h\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\\
&= \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1}\\
&= \lim_{n\to\infty}\frac{\left(1+\frac{x}{n}\right)^{n}}{1+\frac{x}{n}}\\
\end{aligned}$$
Since the limit for numerator and denominator exists independently, we can calculate them separately. The numerator is just the definition of $\mathrm e^x$, and the limit of the denominator is $1$, so we arrive at
$$\frac{\mathrm d}{\mathrm dx}\mathrm e^x = \mathrm e^x$$
A: If the definition of $e^x$ is "the differentiable solution to $f(x+y) = f(x)f(y)$ with $f'(0) =1$, this way works:
Putting $y = 0$, $f(x) = f(x)f(0)$ for all $x$, so $f(0) = 1$.
$(f(x+h)-f(x))/h = (f(x)f(h)-f(x))/h
= f(x)(f(h)-1)/h
= f(x)(f(h)-f(0))/h
$.
Taking the limit as $h \to 0$, $f'(x) = f'(0)f(x)$.
We now can use the differential equation approach.
Note: If this seems familiar, I have used this answer previously
in a similar context.
A: Define $e$ implicitly by $\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1$. Calculate,
$$ \frac{d}{dx} e^x = \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^{x}\lim_{h \rightarrow 0} \frac{e^h-1}{h} = e^x.$$
This definition assumes that properties of exponential functions are somehow known. 
In contrast, the definition that defines the $\ln(x) = \int_{1}^{x} \frac{dt}{t}$ allows you to derive properties of the natural log. Then the exponential function is introduced as the inverse function and its properties can be induced from those already proven for the natural log.
Logically the definition of the natural log as primary has advantages. But, pedagogically if you wish to discuss the exponential function before integral calculus then some sort of chicanery is required.
A: We have $e^x$ = $(logx)^{-1}$ 
Also we know $(f^{-1}(x))^{'}$ = $\frac{1}{f^{'}(f^{-1}(x))}$
$f^{'}(x)$ = $\frac{1}{x}$
Thus, $(e^{x})^{'}$ = $((logx)^{-1})^{'}$ = $\frac{1}{\frac{1}{(logx)^{-1}}}$ = $(logx)^{-1}$  = $e^x$
