Derivative of $e^x$ (ii)

Question

This may be quite basic. I'm trying to understand the idea that $\frac{d(e^x)}{dx}=e^x$. Could someone please explain to me what's wrong with the following argument?

$e$ can be defined as the positive number such that

$$\frac{d(e^x)}{dx}=e^x\;\;\;\;\;(1)$$

Suppose we apply differential triangles with unit bases, like the ones pictured below.

Since $dx=1$,

$$\frac{d(e^x)}{dx}=d(e^x)\;\;\;\;\;(2)$$

so from $(1)$ and $(2)$ we have

$$d(e^x)=e^x\;\;\;\;\;(3)$$

Now, by considering the verticals of the triangles it seems to me that

$$d(e^x)=e^{x_{n+1}}-e^{x_n}\;\;\;\;\;(4)$$

so from $(3)$ and $(4)$ we have

$$e^x=e^{x_{n+1}}-e^{x_n}\;\;\;\;\;(5)$$

Evaluating at $n=1$,

$$e^x=e^{x_2}-e^{x_1}\\\;\;\;\;\;\;\;\;\;=e^2-e^1\;\;\;\;\;(6)$$

but evaluating at $n=2$,

$$e^x=e^{x_3}-e^{x_2}\\\;\;\;\;\;\;\;\;\;\;\;=e^3-e^2\;\;\;\;\;(7)$$

so something seems to be wrong.

Answer 1

The derivative operator is $\frac{d}{dx}$ and I write the derivative of a function as $\frac{d}{dx}f$ for clarity (notice that I keep $\frac{d}{dx}$ together as 1 object with $f$ off to the side).

The differential operator is $d$ and the differential of a function $f$ is written $df$. The definition is $df := \frac{d}{dx}f \; \Delta x$. For a given function $f$, notice how $\frac{d}{dx}f$ itself is a function of $x$. For a given function $f$, notice how $df$ is itself a function of $x$ (from the $\frac{d}{dx}f$ part) and $\Delta x$. So if you give me an $x$ value, and an increment value $\Delta x$, then I can plug those numbers into $df (x, \Delta x)$ to get the value of the change in height of the tangent line. $\frac{d}{dx}f$ gives the slope, but $df$ gives the change in height of a tangent line.

[Given an independent variable, the increment $\Delta x$ is usually written as $dx$. If you wanted to be really rigorous in your thinking however, then you would remember that you can only take differentials (or derivatives) of functions. So the $x$ here in $dx$ is actually a function. Let me explain. Consider $f: x \mapsto e^x$. You can either write $df$ or $d(e^x)$. Now consider the identity function $g: x\mapsto x$. The differential would be written $dg$ or $d(x)$ where the $x$ in $d(x)$ is that 2nd $x$ after the mapsto arrow $\mapsto$...just as $e^x$ is the thing following the mapsto arrow in $x \mapsto e^x$. Okay so we know what $dx$ now means. But why can we equate $\Delta x$ with $dx$? We just follow the defintion. The derivative of the identity function is $1$ so $d(x) = 1 \Delta x$ or just $d(x) = \Delta x$ or just $dx = \Delta x$]

In your question, following the definition of a differential, we jump straight to step $3$, $d(e^x) = \frac{d}{dx}e^x \Delta x$. Let our increment be $1$, so $d(e^x) = e^x$

At $x = 0$, the $x = 0$ tangent line rises by $d(e^x)|_{x=0, \Delta x = 1} = e^0$

At $x = 1$, the $x = 1$ tangent line rises by $d(e^x)|_{x=1, \Delta x = 1} = e^1$

At $x = 2$, the $x = 2$ tangent line rises by $d(e^x)|_{x=2, \Delta x = 1} = e^2$

Answer 2

It follows from the functional equation for $f(x) = e^x$: $f(x+y) = f(x)f(y)$ and the assumption that $f(x)$ is differentiable at $x=0$.

Putting $y=0$, $f(x) = f(x)f(0)$ so, if $f$ is not zero at some $x$, $f(0) = 1$.

$\begin{array}\\ f'(x) &=\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\\ &=\lim_{h \to 0}\dfrac{f(x)f(h)-f(x)}{h}\\ &=\lim_{h \to 0}f(x)\dfrac{f(h)-1}{h}\\ &=f(x)\lim_{h \to 0}\dfrac{f(h)-1}{h}\\ &=f(x)\lim_{h \to 0}\dfrac{f(h)-f(0)}{h}\\ &=f(x)f'(0)\\ \end{array} $

So any differentiable solution to $f(x+y)=f(x)f(y)$ satisfies $f'(x) = f(x)f'(0)$ and $e^x$ is the particular one for which $f'(0) = 1$.

Note that all differentiable solutions to $f(x+y)=f(x)f(y)$ are of the form $f(x) = a^x$ so that $f'(x) = \ln(a)a^x =\ln(a)f(x) $ so that $f'(0) = \ln(a) $.