Why does the derivative of an exponential function depend inherently upon the natural logarithm

calculusexponential functionlimitslogarithms

So in differentiating any exponential function we find a common pattern. Namely, the derivative is equal to the function itself multiplied by a constant which does not depend on the input variable. How would one go about computing this constant? I guess I'm looking to understand how we deduce that this constant is equal to the natural logarithm of the exponential's base?
\begin{align}
\frac{d}{dx}2^x
&=\lim_{h\to0}\frac{2^{x+h}-2^x}{h}\\
&=2^x\lim_{h\to0}\frac{2^h-1}{h}\\[6px]
&=2^x\ln(2)
\end{align}

Best Answer

Your question assumes that you have some definition of $2^x$ which does not use logarithm. The problem of defining $a^x$ for $a>0,x\in\mathbb {R} $ without the use of logarithms is difficult and most textbooks avoid this approach.

Anyway assuming that this difficult part is taken care of, one can proceed as in your question and get $$(a^x) '=a^x\lim_{h\to 0}\frac{a^h-1}{h}\tag{1}$$ From here we have to embark on the journey of exploring the limit $$\lim_{h\to 0}\frac{a^h-1}{h}\tag{2}$$ by proving the following:

  • the limit $(2)$ exists if $a>0$ and hence defines a function $F:\mathbb{R}^{+} \to\mathbb {R} $ via $F(a) =\lim\limits_{h\to 0}\dfrac{a^h-1}{h}$.
  • $F$ is strictly increasing and continuous with $F(1)=0$.
  • $F(ab) =F(a) +F(b), a>0<b$.
  • $F(a^b) =bF(a), a>0,b\in\mathbb {R} $.
  • $F$ is differentiable with derivative $F'(x) =1/x$.
  • There is a positive number $e>1$ such that $F(e) =1$. And we have $F(e^x) =x, x\in\mathbb {R} $ and $e^{F(x)} =x, x\in\mathbb {R} ^{+} $. Thus $F(x) $ and $e^ x$ are inverses to each other.

And once we have proved the above we can just say that $F(x) $ is the natural logarithm of $x$.

All of the above including the difficult definition of $a^x$ has been provided with details in my blog post.

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