Derivative of exponential functions Can anyone present an intuitive reason for why the derivatives of exponential functions, lets say, as apposed to polynomials, grow more rapidly than the functions themselves? 
i.e.
$$
y = e^{x^2}\\
\frac{\mathrm{d}y}{\mathrm{d}x} = 2 x e^{x^2}
$$
I would appreciate an answer that does not simply go out and show algebraic manipulations (of limits etc.) which lead to the desired result. I am much more interested in a, at least partially, verbal explanation. 
Thank you! :)
 A: You must consider that there are some exponential functions such as $1.00001^x$ that clearly grows much slower than itself and there are functions such as $(100^{100})^x$ or even $10^x$ that clearly grow faster than itself. This can be determined by looking at a graph or by doing some numerical calculations.
Now consider the derivative of $a^x$. This is equal to $$\lim_{h\to 0} \dfrac{a^{h+x}-a^x}{h} .$$ Use exponent rules and factor out an $a^x$ to find that $$\lim_{h\to 0} \dfrac{a^{h+x}-a^x}{h} = a^x \lim_{h\to 0}\dfrac{a^h - 1}{h}.$$
Notice that $a^x$ is the function itself and $\lim_{h\to 0}\dfrac{a^h - 1}{h}$ is simply a constant (which happens to be equal to $\log(a)$). This means exponential functions grow some constant multiple of themselves and, if this constant is greater than $1$, they will grow faster than the function itself.
Often times people define $e$ to be the value of $a$ such that $$\lim_{h\to 0}\dfrac{a^h - 1}{h} = 1.$$
And $\log(x) = \log_e(x)$. The existence of the value $e$ can be justified because one can graphically determine that $1.00001^x$ grows slower than itself and $10^x$ grows faster as mentioned before. That means there should be an "$e$", $0 \lt e \lt 10$.
Polynomial functions can grow faster than themselves on an interval but as $x \to \infty$ the polynomial with the higher degree will be larger in magnitude for any polynomial. This is why this result does not hold for polynomials as well; the derivative of a polynomial has a degree of one less than the polynomial itself.
A: One possible explanation lies in the definition of $\mathrm e^x$: it is the sum of the power series:
$$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots +\frac{x^k}{k!}+\dotsm $$
which converges for all $x$ and can be differentiated term by term. Now you can easily check that
$$\frac{\mathrm d}{\mathrm d\mkern1mux}\biggl(\frac{x^k}{k!}\biggr)=\frac{x^{k-1}}{(k-1)!},$$
so that
$$\frac{\mathrm d}{\mathrm d\mkern1mu x}\Bigl(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsm \biggr)=1+x+\frac{x^2}{2!}+\dotsm $$
A: My intuition is as follows. When you have a monotonic function $f(x)$ and you change it to $f(g(x))$ (where $g(x)>x$) then the value increases, because instead of using $x$ now you go further down the number line and look up the value of $f$ at point $g(x)$ instead of at $x$.
The same source of growth (looking up values further down the number line) is available for the derivative, but the derivative also has an additional source of growth, namely that all this looking-further-and-further causes the slope of the resulting function to be "horizontally squeezed" (increased). Concretely, the derivative acquires an extra factor corresponding to the speed at which the how-far-we're-looking function changes, i.e. the derivative of the inner function. This effect is usually called the "chain rule".
