Derivative Rules via Dual Numbers

This post is a tidbit that I wrote on a whim.

Let \(\varepsilon_1\) be an indeterminate such that \(\varepsilon_1^2=0\).

Then for an analytic function \(f\), \(f(a+b\varepsilon_1)=f(a)+f'(a)b\varepsilon_1\) for some \(a,b\in\mathbb{R}\), using the Taylor series of \(f\) centred at \(a\).

Equivalence of dual numbers

Two dual numbers \(u=a+b\varepsilon_1\) and \(v=c+d\varepsilon_1\) are equal iff \(a=c\) and \(b=d\) for \(a,b,c,d\in\mathbb{R}\).

Proof: \(a+b\varepsilon_1=c+\varepsilon_1\implies (a-c)=(d-b)\varepsilon_1\). Multiplying both sides by \(\varepsilon_1\) gives \((a-c)\varepsilon_1=(d-b)\varepsilon_1^2=0\). Since \(\varepsilon_1\) is non-zero by definition, we have that \(a-c=0\implies a=c\). It follows, therefore, that \(b=d\) by the same reasoning.

(Note that the reverse is fairly trivial; if \(u=a+b\varepsilon_1\) and \(a=c\) and \(b=d\), then \(u=c+d\varepsilon_1=v\).)

Chain Rule

Let \(F(x)=f(g(x))\); then \(F(a+b\varepsilon_1)=F(a)+F'(a)b\varepsilon_1\).

On the other hand, \(f(g(a+b\varepsilon_1))=f(g(a)+g'(a)b\varepsilon_1)=f(g(a))+f'(g(a))g'(a)b\varepsilon_1\).

Equating the above equations, we have that:

\[ F(a)+F'(a)b\varepsilon_1 = f(g(a))+f'(g(a))g'(a)b\varepsilon_1\]

By Equivalence of Dual Numbers, we have that:

\[ F(a) = f(g(a))\]

\[ F'(a)b=f'(g(a))g'(a)b\implies F'(a)=f'(g(a))g'(a)\]

Therefore:

\[ \frac{\text{d}}{\text{d}x}f(g(x))=f'(g(x))g'(x)\]

Product Rule

Let \(F(x)=f(x)g(x)\); then \(F(a+b\varepsilon_1)=F(a)+F'(a)b\varepsilon_1\).

On the other hand, \(f(a+b\varepsilon_1)g(a+b\varepsilon_1)=(f(a)+f'(a)b\varepsilon_1)(g(a)+g'(a)b\varepsilon_1)\). Expanding and factoring into real and non-real parts, we get \(f(a+b\varepsilon_1)g(a+b\varepsilon_1)=f(a)g(a)+(f(a)g'(a)+f'(a)g(a))\varepsilon_1\).

Equating the above equations once again, we get:

\[ F(a)+F'(a)b\varepsilon_1 = f(a)g(a)+(f(a)g'(a)+f'(a)g(a))\varepsilon_1\]

By Equivalence of Dual Numbers, we have that:

\[ F(a) = f(a)g(a)\]

\[ F'(a)b = f(a)g'(a)+f'(a)g(a)b\implies F'(a) = f(a)g'(a) = f(a)g'(a)+f'(a)g(a)\]

Therefore:

\[ \frac{\text{d}}{\text{d}x}f(x)g(x)=f(x)g'(x)+f'(x)g(x)\]

Quotient Rule

Let \(F(x)=\frac{f(x)}{g(x)}\); then \(F(a+b\varepsilon_1)=F(a)+F'(a)b\varepsilon_1\).

On the other hand, \(\frac{f(a+b\varepsilon_1)}{g(a+\varepsilon_1)}=\frac{f(a)+f'(a)b\varepsilon_1}{g(a)+g'(a)b\varepsilon_1}\). For non-zero \(g(a)\), we multiply numerator and denominator, and expand, by \(g(a)-g'(a)b\varepsilon_1\), giving us:

\[ \frac{f(a)}{g(a)}+\frac{-f(a)g'(a)+f'(a)g(a)}{g(a)^2}\varepsilon_1\]

Equating the above equations, we have that:

\[ F(a)+F'(a)b\varepsilon_1=\frac{f(a)}{g(a)}+\frac{-f(a)g'(a)+f'(a)g(a)}{g(a)^2}b\varepsilon_1\]

By Equivalence of Dual Numbers, we have that:

\[ F(a) = \frac{f(a)}{g(a)}\]

\[ F'(a)b=\frac{-f(a)g'(a)+f'(a)g(a)}{g(a)^2}b\implies F'(a)=\frac{-f(a)g'(a)+f'(a)g(a)}{g(a)^2}\]

Therefore:

\[ \frac{\text{d}}{\text{d}x}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\]