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It's known the the derivative of exponential function $a^x$ is $xa^{x-1}$.

If I play $e$ as $a$, we'll get $(a^x = \operatorname e^x)' = x \operatorname e^{x-1}$.

Why does $(\operatorname e^x)' = \operatorname e^x$?

jimjim
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Billie
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4 Answers4

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As T. Bongers has pointed out, you don't have the correct rule for exponential functions. The real rule for functions like these is $$ \frac d{dx} a^x = \ln(a)\cdot a^x $$ Once you have this rule, it's clear what should happen for $e^x$. Whether or not this derivation is inherently circular depends on your definition of $e$.

Ben Grossmann
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If you start with $$e^x=1 + x + \frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots$$ and differentiate it term by term you get $$\qquad \qquad \, \, \, 0+ 1 + \frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots = e^x$$

Henry
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If you take your definition of $e^x$ as $\lim_{n\to\infty}(1+\frac{x}{n})^n$

Then differentiating we see that,

$$\frac{d}{dx}(1+\frac{x}{n})^n=\frac{n}{n+x}(1+\frac{x}{n})^n$$ $$\lim_{n\to \infty}\frac{d}{dx}(1+\frac{x}{n})^n=\lim_{n\to\infty}\frac{n}{n+x}(1+\frac{x}{n})^n$$

$$\frac{d}{dx}\lim_{n\to\infty}(1+\frac{x}{n})^n=\lim_{n\to\infty}\frac{n}{n+x}\lim_{n\to\infty}(1+\frac{x}{n})^n$$

$$\frac{d}{dx}e^x=e^x$$

If you're still not satisfied you could justify the interchanging of the limits in the second step.

But generally speaking when you ask why something is true, I think you need to specify what sort of framework were supposed to work in, like what tools, axioms or logics you expect us to use. For many the fact that $\frac{d}{dx}e^x=e^x$ could simply be taken as the definition of $e^x$ up to some constant scaling factor.

Ethan Splaver
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You can certainly change the variable $a$ to the variable $e$ (you can call it whatever you want, how about $math^x$?), but you can't change the meaning of $a$ to the exponential function! You can find the answer you're looking for when going back to the definition of derivatives.

The derivative of the exponential function: \begin{align} \left( \mathrm{e}^{x} \right)' &= \lim_{h \to 0} \frac{ \mathrm{e}^{x+h} - \mathrm{e}^x }{h} \\ &= \lim_{h \to 0} \frac{\mathrm{e}^x \mathrm{e}^h - \mathrm{e}^x}{h} \\ &= \mathrm{e}^x \lim_{h \to 0} \frac{\mathrm{e}^h - 1}{h} = \mathrm{e}^x. \end{align} The limit of the last fraction can be found by expanding the fraction in a Taylor series and taking the limit ($(\mathrm{e}^h-1)/h = 1 + h/2 + O(h^2)$, hence the limit is $1$).

The derivative of the power function: \begin{align} (x^n)' &= \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}\\ &= \lim_{h \to 0} \frac{ \left[ (x+h)-x \right] \left[(x+h)^{n-1} + (x+h)^{n-2}x + \cdots + x^{n-1} \right]}{h} \\&= \lim_{h \to 0} (x+h)^{n-1} + (x+h)^{n-2}x + \cdots + x^{n-1} = n x^{n-1}. \end{align}

The derivative of $a^x$: \begin{align} \left( \mathrm{a}^{x} \right)' &= \lim_{h \to 0} \frac{ \mathrm{a}^{x+h} - \mathrm{a}^x }{h} \\ &= \lim_{h \to 0} \frac{\mathrm{a}^x \mathrm{a}^h - \mathrm{a}^x}{h} \\ &= \mathrm{a}^x \lim_{h \to 0} \frac{\mathrm{a}^h - 1}{h} = \ln(a) \mathrm{a}^x. \end{align} Again, the last limit can be found by Taylor expansion: $$\frac{\mathrm{a}^h - 1}{h} = \ln(a) + \frac{\ln(a)^2 h}{2} + O(h^2).$$

Numbersandsoon
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  • If you had that Taylor series expansion to begin with, you could have just point-wise differentiated each monomial in the power series to show there derivatives were the same, so really it mostly comes out to how the OP is defining $f(x)=e^x$. – Ethan Splaver Nov 27 '13 at 06:39
  • True, what would be another simple way to find those limits without Taylor series expansion? – Numbersandsoon Nov 27 '13 at 06:51