## Proof: e^x Differentiates to Itself

To prove that $$\frac{d}{dx}(e^x)=e^x$$, it is sufficient to prove that if $$\frac{d}{dx}(a^x)=a^x$$ then $$a=e$$. We will use the definition of e as a limit, i.e. that:

$e=\lim_{n \to 0} (1+n)^{\frac{1}{n}}$

## Proof

Let a be a constant such that $$\frac{d}{dx}(a^x)=a^x$$.

For this proof, we'll need to know that $$\frac{d}{dx}(\log_a x) { \Big{|}} _{x=1}=1$$. For simplicity, I'll avoid differentiating $$\log_a x$$. We know that the value of $$a^x$$ at $$x=0$$ is 1, and also that $$\frac{d}{dx}(a^x)=a^x$$, so $$\frac{d}{dx}(a^x) { \Big{|}} _{x=0}=a^0=1$$. We can plot this information on a graph like so:

(*graph)

Now the inverse of a function is its reflection in the line $$y=x$$, so we can plot the graph of $$\log_a x$$ onto our diagram like so:

(*graph)

Now notice that if we reflect the tangent we drew on the $$a^x$$ curve in the line $$y=x$$, we get a tangent to the $$\log_a x$$ curve at the point $$(1,0)$$. Since the initial tangent had a gradient of 1, and we reflected it in $$y=x$$ which also has a gradient of 1 (so they are parallel), the new tangent will also have a gradient of one, and so we have shown that $$\frac{d}{dx}(\log_a x) { \Big{|}} _{x=1}=1$$.

Now we have to show that $$a$$ is indeed $$e=2.71828...$$ as claimed. Consider the derivative of $$\log_a x$$ from first principles:

$\frac{d}{dx}(\log_a x)=\lim_{h \to 0} \frac{\log_a(x+h)-\log_a(x)}{h}$

Now let $$x=1$$. We know that $$\frac{d}{dx}(\log_a x) { \Big{|}} _{x=1}=1$$, so substituting that in gives:

$1=\lim_{h \to 0} \frac{\log_a(1+h)-\log_a(1)}{h}$

We know that $$\log_a(1)=0$$ since $$1=a^0$$, so:

$1=\lim_{h \to 0} \frac{\log_a(1+h)}{h}$

$1=\lim_{h \to 0} \frac{1}{h} \log_a(1+h)$

$1=\lim_{h \to 0} \log_a{\left[(1+h)^{\frac{1}{h}}\right]}$

Now we can do $$a$$ to the power of both sides, like so:

$a^1=a^{\lim_{h \to 0} \log_a{\left[(1+h)^{\frac{1}{h}}\right]}}$

We can now take the limit out of the power (this is because of the algebra of limits and the fact that $$a^x$$ is continuous):

$a=\lim_{h \to 0} a^{\log_a{\left[(1+h)^{\frac{1}{h}}\right]}}$

Now the $$a^{(\cdots)}$$ undoes the $$\log_a$$ on the right side, so:

$a=\lim_{h \to 0} (1+h)^{\frac{1}{h}}$

And this limit does in fact equal $$2.71828...$$, and you can try some very small values of h to confirm this.

$\therefore a=e$

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

$Q.E.D.$