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Is there a way to explain the differentiation of logarithmic and exponential functions without invoking the assumption that e exists? Is the existence of e necessary to differentiate these functions?

charlie_sar
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    Since the derivative of $a^x$ involves the natural logarithm (logarithm with base $e$), I don't see how you can do it without $e$. – Andrei Dec 29 '20 at 22:36
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    Depends on your definition. Some people define $\exp(x)$ as the unique solution to the IVP $y'(x)=y(x)~|~y(0)=1$. Without using this rather synthetic definition, one must show the number $e$ exists. – K.defaoite Dec 29 '20 at 22:42
  • How do you define exponential function $e^x$ without $e$? – herb steinberg Dec 29 '20 at 22:42
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    The existence of $e$ is a fact, not an assumption. It's not clear what you mean. – Gerry Myerson Dec 29 '20 at 22:55
  • @herbsteinberg: Most of the common definitions of the exponential function don't use the number $\mathrm e$. In fact, $\mathrm e$ is usually defined as $\exp(1)$, after the exponential function has been defined. Be that via the IVP it satisfies, or as a power series, or via its functional equation, or some other definition. – Vercassivelaunos Dec 29 '20 at 22:56
  • Did my answer help you? Or is there something you are unsure about? – Joe Jan 23 '21 at 01:37

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If we define $\log(x)$ as $\int_{1}^{x}\frac{1}{t} \, dt$, then once we have established that $\log$ is strictly increasing, $\exp$ can be defined as its inverse. Furthermore, $\log'(x)=1/x$ by the first part of the fundamental theorem of calculus. We can then use the formula for the derivative of an inverse function, which can be derived in the following way: \begin{align} y &= f^{-1}(x) \\ x &= f(y) \\ \frac{dx}{dy} &= f'(y) \\ \frac{dy}{dx} &= \frac{1}{f'\left(f^{-1}(x)\right)} \, . \end{align} Hence, \begin{align} \exp'(x) &= (\log^{-1})'(x) \\ &= \frac{1}{\frac{1}{\exp(x)}} \\ &= \exp(x) \, . \end{align} This approach makes no reference to the constant $e$, which is simply the value of $\exp(1)$. Indeed, one could argue that it is not $e$ which is of fundamental importance, but rather the exponential function.

All this being said, full appreciation of the exponential function can only be reached once we have established that $\exp(x+y)=\exp(x) \cdot \exp(y)$. This provides the motivation behind using the more familiar notation $e^x$ instead of $\exp(x)$. $a^x$ can then be defined as $e^{x\log(a)}$ for $a>0$.

Joe
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