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! :)
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.