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.

\[\therefore a=e\]

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

\[Q.E.D.\]