Derivatives of $\sin x$ and $\exp x$ using differentials / dual numbers I want to introduce a concept of a differential $dx$ to my students and derive all the basic derivatives using it. Now, I define the differential to satisfy $dx \neq 0$, but $(dx)^2 = 0$. Therefore, it is an example of a dual number. Is there a way to prove that $$\exp (dx) = 1 + dx, \quad \sin(dx) = dx, \quad \cos(dx) = 1,$$ just by using the defining property of the differential and not the series expansion (which needs the derivatives I am trying to derive) of the above functions?
 A: Basically, if you take a geometric proof from a calculus textbook that the limit of $\lim_{x\to0}\frac{\sin x}{x}=1$, I suppose you could rephrase that and wave your hands a little to argue that $\frac{\sin(\mathrm dx)}{\mathrm dx}=1$, which gives $\sin(\mathrm dx)=\mathrm dx$. Then you can use the conjugate trick to get $\frac{1-\cos(\mathrm dx)}{\mathrm dx}=0$ (whence $\cos(\mathrm dx)=1$). Then you can prove the derivative of sine in a usual way. Similarly, if you define $\exp$ so that $\lim_{h\to0}\frac{\exp h-1}{h}=1$, then you can handwave and say that $\frac{\exp\left(\mathrm dx\right)-1}{\mathrm dx}=1$ so that $\exp(\mathrm dx)=1+\mathrm dx$.
Except there's a problem with all of that: you can't actually divide by $\mathrm dx$ in the dual numbers. You really have to do something like develop the infinite series and say "we'll use these series to define $\sin$, $\cos$, and $\exp$ on the dual numbers" if you want to do things formally. And if you don't want to do things formally, I'd recommend handwaving at the hyperreals over handwaving at the dual numbers or something.
