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

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}

• How do you define $2^x$ for $x$ being real ? Commented Mar 25, 2019 at 20:20
• Comment: that's why $e$ is a special number Commented Mar 25, 2019 at 20:34
• Do you know the Binomial series? Commented Mar 25, 2019 at 20:36

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 $$F(1)=0$$.
• $$F(ab) =F(a) +F(b), a>0.
• $$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.

Let $$h=\dfrac1n$$. We have

$$\frac{2^h-1}h=a\iff(1+ah)^{1/h}=\left(1+\frac an\right)^n=2$$

Taking the limit for $$h\to0$$,

$$\ln 2=a\iff e^a=2.$$

The constant $$e$$ appears as the limit of

$$\left(1+\frac 1n\right)^n.$$

Notice there is always a constant, call it $$C_b$$, where $$\frac {db^x}{x} = C_b\cdot b^x$$. We also notice that if $$0 < b < c$$ then $$C_b < C_c$$. And it appears that if we consider $$C_x$$ as a function of $$x$$ thatn $$C_x$$ is continuous, 1-1, and increasing. As $$C_2 < 1$$ and $$C_3 > 1$$ it seems there must by some number $$E$$ so that $$C_E = 1$$.

This means that as $$b^x = E^{x\log_E b}$$ that by the chain rule $$\frac {db^x}{dx} = (C_E\cdot E^{x\log_E b})\cdot\log_E b = \log_E b\cdot b^x$$. But $$\frac {db^x}{dx} = C_bb^x$$. So that means $$C_b = \log_E (b)$$.

So we can use that as our definition. $$e$$ is the base where $$\frac {de^x}{dx}=e^x$$. And as a result $$\frac {db^x}{dx} =\frac {de^{x\log_e b}}{dx} =\log_e b\cdot b^x$$.

All neat clean and accurate.

... but backwards.

Thing is .... we haven't actually defined what $$b^x$$ means if $$x$$ is irrational.

So most texts define $$e$$ and $$\ln x$$ in other ways and derive these results.

Write $$k=\ln a$$ so $$\frac{d}{dx}a^x=\frac{d}{dx}e^{kx}=ke^{kx}$$ by the chain rule. This is $$a^x\ln a$$.