# 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 dy/dx=2xe^x^2 I am interested in a verbal explanation.

Jairo Decker 2022-10-12 Answered
Why the derivatives of exponential functions, lets say, as apposed to polynomials, grow more rapidly than the functions themselves?
i.e.
$y={e}^{{x}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}y}{\mathrm{d}x}=2x{e}^{{x}^{2}}$
I am interested in a verbal explanation.
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## Answers (2)

Szulikto
Answered 2022-10-13 Author has 22 answers
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 $\left({100}^{100}{\right)}^{x}$ 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
$\underset{h\to 0}{lim}\frac{{a}^{h+x}-{a}^{x}}{h}.$
Use exponent rules and factor out an ax to find that
$\underset{h\to 0}{lim}\frac{{a}^{h+x}-{a}^{x}}{h}={a}^{x}\underset{h\to 0}{lim}\frac{{a}^{h}-1}{h}.$
Notice that ${a}^{x}$ is the function itself and $\underset{h\to 0}{lim}\frac{{a}^{h}-1}{h}$ 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
$\underset{h\to 0}{lim}\frac{{a}^{h}-1}{h}=1.$
And $\mathrm{log}\left(x\right)={\mathrm{log}}_{e}\left(x\right)$. 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.

Polynomial functions can grow faster than themselves on an interval but as $x\to \mathrm{\infty }$ 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.
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Winston Todd
Answered 2022-10-14 Author has 3 answers
One possible explanation lies in the definition of ${\mathrm{e}}^{x}$: it is the sum of the power series:
$1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\cdots +\frac{{x}^{k}}{k!}+\cdots$
which converges for all x and can be differentiated term by term. Now you can easily check that
$\frac{\mathrm{d}}{\mathrm{d}\phantom{\rule{0.056em}{0ex}}x}\left(\frac{{x}^{k}}{k!}\right)=\frac{{x}^{k-1}}{\left(k-1\right)!},$
so that
$\frac{\mathrm{d}}{\mathrm{d}\phantom{\rule{0.056em}{0ex}}x}\left(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\cdots \right)=1+x+\frac{{x}^{2}}{2!}+\cdots$
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