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I don't understand the reasoning behind it. I was trying to figure it out with the chain rule in the case of $e^x$, but the I would get $(X(e)^x-1)(e^x)$ and I can't find the reasoning by my own

mjw
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user841190
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    If the derivative is with respect to $x$, then $x$ is in the exponent. Please don't confuse this with the derivative of $x^n$. – mjw Oct 23 '20 at 18:10
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    You say something rather alarming when you wrote (X(e)^x-1)(e^x)... it looks as though you are remembering how $\dfrac{d}{dx}[x^n] = n\cdot x^{n-1}$. That only has to do with when $x$ is the base... not when $x$ is the exponent. – JMoravitz Oct 23 '20 at 18:10
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    $$ \frac{d}{{dx}}a^x = \frac{d}{{dx}}e^{x\log a} = e^{x\log a} \frac{d}{{dx}}(x\log a) = e^{x\log a} \log a = a^x \log a $$ – Gary Oct 23 '20 at 18:11
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    As for "why is the derivative of $e^x$ equal to $e^x$"... to answer that we must first ask you how do you have $e$ defined? How do you have $e^x$ defined? This should in most circumstances be a matter of definition – JMoravitz Oct 23 '20 at 18:11
  • Welcome to MSE. If you need help formatting math on this site, here's a tutorial To start with, surround math expressions by $ signs. – saulspatz Oct 23 '20 at 18:16
  • There's a lot of info at the duplicate target. But if you still need help with this after reading all those answers, then you can edit this question and try to explain what you need, and answer the questions JMoravitz asked you. There are several different ways to approach this topic, and it's hard for us to help you without knowing stuff like how you have defined $e$. – PM 2Ring Oct 23 '20 at 19:12

3 Answers3

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These are different!

$$\frac{d}{dx} x^n = n x^{n-1}, \quad \text{ Here }x\text{ is being raised to a power}.$$

$$\frac{d}{dx} e^x = e^x, \quad \text{ Here }x\text{ is the exponent}.$$ Let's show this: $$\frac{d}{dx} e^x = \lim_{h\to0} \frac{e^{x+h}-e^x}{h}= \lim_{h\to0} \frac{(e^{h}-1)}{h}e^x$$

Now we need a definition for $e^x$. Here is one:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+ \cdots$$

Now $$\lim_{h\to0} \frac{e^h-1}{h} = \lim_{h\to0}1+h+\frac{h^2}{2}+\cdots=\lim_{h\to0}e^h = 1,$$ so $$\frac{d}{dx} e^x =e^x.$$

mjw
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Perhaps unsatisfyingly, $f(x) = e^x$ is sometimes defined as the unique function such that $$ \frac{d}{dx}f(x) = f(x) $$ with $f(0) = 1$.

From there, we have $$ a^x = (e^{\ln a})^x = e^{x \ln a} $$ so that $$ \frac{d}{dx} a^x = \frac{d}{dx} e^{x \ln a} = e^{x \ln a} \frac{d}{dx} (x \ln a) = (\ln a) e^{x\ln a} = (\ln a) a^x $$ by the chain rule.

Charles Hudgins
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Derive $e^x$ and $x^n$ from first principles to see why they are what they are.

$$f'(x) = \lim\limits_{h \to 0} {f(x+h) - f(x) \over h}$$

vvg
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