On The Differential Of exp(x)



I take an undergrad class in maths for chemists without A-level maths. The aim of the class is to take someone with GCSE knowledge and get them ready for some of the quantum theory that they’ll be running into during their degree. Now this is a tall order, but it has proved successful in the past. This articles is aimed at someone in their shoes.

Anyway, we were dealing with the differential of \( e^{x}\) a few months ago and I realised that I don’t think I’ve ever really worried myself too much of the proof of this. A common ‘proof’ that one encounters on the internet relies on using a Taylor expansion about zero (don’t worry if you haven’t met them):

\[exp(x)\approx 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\ldots\]

Which when differentiated clearly gives itself. BUT the construction of a Taylor expansion relies on you knowing the differential of the function that you are trying to approximate! So this is a circular argument.

Lets go back to basics – what do we mean by a differential? Well, we’re talking about the gradient of a function, or the rate of change. Think about that long hill on the way back from school/work – the steepness of that hill is its gradient. So, to take the analogy further, it’s how much do we go up for every step we go forward.

Lets say that the change in height is \(\Delta y \) after a step of \(\Delta x\). Then our gradient is \(\frac{\Delta y}{\Delta x} \) right? Now, this is all well and good, but what would be really useful would be to know the gradient at exactly one point – how much do we increase in height on this hill with a tiny, nay, infinitesimally small step? Let’s write this as:

\[\lim_{x\rightarrow 0}\frac{\Delta y}{\Delta x}=\frac{\mathrm{d} y}{\mathrm{d} x}\]

Which is just a shorthand way of writing what we said before. In the limit (\(\lim\)) that \( \Delta x \) becomes infinitesimally small (approaches zero) we find the derivative. So Lets take an example of \(f(x)=2\).

\[\begin{aligned} \frac{\mathrm{d} y}{\mathrm{d} x} & = \lim_{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x} \\ & =  \lim_{\Delta x\rightarrow 0} \frac{\Delta f(x)}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{(x+\Delta x)^{2} - x^{2}}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{x^{2}+2\Delta x + (\Delta x)^{2} - x^{2}}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{2\Delta x + (\Delta x)^{2}}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} 2x + \Delta x \\ & = 2x + 0 = 2x \end{aligned}\]

Follow what’s happening? \(\Delta y\) is value of the function at the height we’ve climbed to (\(f(x+\Delta x)\) ) minus the height we started at (\( f(x)\) ). We can’t set \(\Delta x\)  to zero straight away, otherwise we’ll divide by zero and all hell will break loose. Instead we expand out our \( \Delta y\) and divide through by \(\Delta x\) . This has the end result that when we set \( \Delta x\) to zero, we have a result of \(2x\) .

So how do we apply this to \(e^{x}\) . And what do we even mean by the exponential function? Well, exponential functions come up quite naturally when we talk about growth or decline of something (money, population, radioactive decay). Imagine you have a back account with some amount of money in (unlike mine!), denoted \(m\) , that receives (compounding) interest at a rate \(\beta\) per year. In one month, you’d receive \(m(1+\frac{\beta}{12})\). If this interest is paid monthly, then in a year you will have in your account:


Don’t forget that this interest is compounding. If it’s paid daily you will have:


Where \(m_{f}\) is the amount of money in your account after a year and \(m_{i}\) is the initial seed capital. See a pattern here? What happens if we keep on increasing the number of divisions in a year up to infinity?

\[m_{f}= \lim_{n\rightarrow\infty} m_{i}(1+\frac{\beta}{n})^{n}\]

This limit situation relates directly to our exponential function. We define \(e^{x}\) as:


Which is the same as:

\[e^{x}=\lim_{k\rightarrow 0}(1+kx)^{\frac{1}{k}}\]

So, if banks paid their interest continuously rather than annually, monthly, daily or whatever, we would get exactly this number. In reality, banks don’t pay like this and populations of bacteria don’t divide continuously. But it is a fantastic model of these processes and we can do loads and loads with this function if we use it, so hence we use it!

We’re nearly ready to answer the original question! Why is \(\frac{\mathrm{d}e^{x}}{dx}=e^{x}\) ?

The easiest way to prove this is by using logarithm functions. If you don’t know about these, they are really simple. If you have an equation like \(a^{b}=c\) then we define the logarithm of \(c\) , to the base \(a\) , as \(b\) . In other words, if we have a number \(c\) that we want to show as \(a^{b}\) , what does \(b\) need to be if we already know \(a\) . We write this as \(\log_{a}\) , or just \(\log\) if the base doesn’t matter. See, told you it was simple. In the same way as there are some simple rules for how we add exponents, there are equivalent rules for logarithms:

 Indicies Logarithms
\((a^{i})^{j}=a^{ij}\) \(i.log(j)=log(j^{j})\)
\(\frac{a^{i}}{a^{j}} = a^{i-j}\) \(log(i)-log(j)=log(i/j)\)
\(a^{i}.a^{j} = a^{i+j}\)  \(log(i)+log(j)=log(ij)\)

OK, we’re all set. Lets calculate the differential of \(\log_{e}\) , which is used so often that we call it \(\ln\) .

\[\begin{align}\frac{\mathrm{d}\ln(x)}{\mathrm{d}x} & = \lim_{\Delta x\rightarrow 0} \frac{\Delta \ln(x)}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{\ln(x+\Delta x)-\ln(x)}{\Delta x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{\ln(\frac{x+\Delta x}{x})}{x} \\ & = \lim_{\Delta x\rightarrow 0} \frac{1}{\Delta x}.\ln(1 + \frac{\Delta x}{x}) \\ & = \lim_{\Delta x\rightarrow 0}.\ln(1 + \frac{\Delta x}{x})^{\frac{1}{\Delta x}}\end{align} \]

Notice anything familiar in the final line? It’s our friend the exponential function, just with an x rather than a \triangle x (also note that we’re dealing with \(\frac{1}{x}\) now, not just \(x\) ).

\[\frac{\mathrm{d}\ln(x)}{\mathrm{d}x} = \ln(e^{\frac{1}{x}})=\frac{1}{x}\]

As \(\ln{a}\) is defined as \(\log_{e}(a)\), then if \(a\) is a simple power of e , like \(e^{2}\) , then \(\ln\left(e^{2}\right)=2\) , as it’s like asking “what power does \(e\) need to be raised to to produced \(e^{2}\) ” – obviously it’s \(2\) !

With this information, lets move on and look at \(\ln(e^{x})\) :


OK, that wasn’t so informative. But if we remember this result, but try to use the chain rule[1] to get here (we’re also going to make the substitution \(u=e^{x}\) ):

\[\begin{align} \frac{d\ln(e^{x})}{dx} & =\frac{d\ln(u)}{du}.\frac{de^{x}}{dx}=1 \\ & =\frac{1}{u}.\frac{de^{x}}{dx}=1\\ & =\frac{1}{e^{x}}.\frac{de^{x}}{dx}=1\end{align}\]

Which means:


BLAM! Hope it wasn’t too dry this week. I got something really interesting about human evolution and it’s consequences for language next time.

[1] If you haven’t encountered this, it’s just a way of dealing with differentiating something which is a function of a function (like \(\ln{e^{x}}\) – \(e^{x}\) is a function \(f(x)\) and it in turn is being acted upon by \(\ln\) ). What we do is differentiate the overall function by the smaller function inside and multiply this by the internal derivative, so in this case \(\frac{d\ln(e^{x})}{dx}=\frac{d\ln(e^{x})}{de^{x}}.\frac{de^{x}}{dx}\).

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