Proof of Reciprocal Rule for One-Dimensional Derivatives

Absolute values are those of that definition, with the properties collected in Properties of the Absolute Value in an Ordered Field; the order arithmetic used below is that of Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field, and $2$ denotes $1+1$.

Proof of claim 1. Write $a=g(x_{0})$, $L=g'(x_{0})$ and $A=|a|$. Since $a\ne 0$, claim 1 of Properties of the Absolute Value in an Ordered Field gives $0\le A$ and $A\ne 0$, hence $0<A$; in particular $a^{-1}$ and $A^{-1}$ exist and are positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. Also $|1|=1$: by claim 1 of the absolute value lemma $|1|$ is $1$ or $-1$, and $|1|=-1$ would give $0\le -1$ and hence $1\le 0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, contradicting claim 6 there. Consequently, for every $w\ne 0$, multiplicativity (claim 4 of the absolute value lemma) gives $|w|\,|w^{-1}|=|w\,w^{-1}|=|1|=1$, so $|w^{-1}|=|w|^{-1}$.

For $h\in\mathbb{R}$ with $h\ne 0$ and $x_{0}+h\in I$ put $b_{h}=g(x_{0}+h)$ and

$$Q(h)=\frac{g(x_{0}+h)-g(x_{0})}{h},\qquad R(h)=\frac{(1/g)(x_{0}+h)-(1/g)(x_{0})}{h}.$$

By hypothesis $b_{h}\ne 0$, and $b_{h}-a=h\,Q(h)$.

Step 1 (an algebraic identity). Expanding, $a^{-1}b_{h}^{-1}(a-b_{h})=b_{h}^{-1}-a^{-1}$, so

$$R(h)=h^{-1}\bigl(b_{h}^{-1}-a^{-1}\bigr)=-a^{-1}b_{h}^{-1}\,\frac{b_{h}-a}{h}=-a^{-1}b_{h}^{-1}Q(h).$$

Since $a^{-1}b_{h}^{-1}\cdot b_{h}a^{-1}L=L\bigl(a^{-1}\bigr)^{2}$, it follows that

$$R(h)+L\bigl(a^{-1}\bigr)^{2}=a^{-1}b_{h}^{-1}\Bigl(-Q(h)+b_{h}a^{-1}L\Bigr),$$

and a direct expansion gives

$$-Q(h)+b_{h}a^{-1}L=-\bigl(Q(h)-L\bigr)+a^{-1}\,(b_{h}-a)\,L.$$

Hence, by the triangle inequality and multiplicativity of the absolute value (claims 5 and 4 of Properties of the Absolute Value in an Ordered Field) together with claim 5 of Elementary Arithmetic in an Ordered Field,

$$\bigl|R(h)+L(a^{-1})^{2}\bigr|\le A^{-1}|b_{h}|^{-1}\Bigl(\bigl|Q(h)-L\bigr|+A^{-1}\bigl|b_{h}-a\bigr|\,|L|\Bigr).$$

We refer to this as the basic estimate.

Step 2 (a crude bound on $b_{h}$). Applying Derivative at an Interior Point to $g$ with $\varepsilon=1$, there is $\delta_{1}>0$ such that $|Q(h)-L|<1$ whenever $0<|h|<\delta_{1}$ and $x_{0}+h\in I$. For such $h$, claim 5 of Properties of the Absolute Value in an Ordered Field gives

$$|Q(h)|=\bigl|\bigl(Q(h)-L\bigr)+L\bigr|\le|Q(h)-L|+|L|<1+|L|.$$

Put $\kappa=1+|L|$; then $0<\kappa$ by claims 6 and 3 of Elementary Order Arithmetic in an Ordered Field and claim 1 of Properties of the Absolute Value in an Ordered Field. Since $b_{h}-a=hQ(h)$, multiplicativity and claim 10 of Elementary Order Arithmetic in an Ordered Field give

$$|b_{h}-a|=|h|\,|Q(h)|<|h|\,\kappa .$$

Step 3 (choice of $\delta$). Let $\varepsilon>0$ be given. Put

$$C=\bigl(2A^{-1}\bigr)A^{-1}\bigl(1+A^{-1}|L|\bigr),$$

which is positive by claims 5, 6, 7 and 8 of Elementary Order Arithmetic in an Ordered Field and claim 2 of Elementary Arithmetic in an Ordered Field, and put $\eta=\varepsilon C^{-1}$, so that $0<\eta$ and $\eta C=\varepsilon$. Let $\mu$ be the smaller of $\eta$ and $A\cdot 2^{-1}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field); then $0<\mu$ by claim 8 there. Now choose $\delta_{1}>0$ as in Step 2; put $\delta_{2}=\mu\kappa^{-1}>0$, so that $\delta_{2}\kappa=\mu$; and choose $\delta_{3}>0$ such that $|Q(h)-L|<\eta$ whenever $0<|h|<\delta_{3}$ and $x_{0}+h\in I$, which exists by the differentiability of $g$ at $x_{0}$. Let $\delta>0$ be the smallest of $\delta_{1},\delta_{2},\delta_{3}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field, used twice).

Step 4 (the estimate). Let $h\in\mathbb{R}$ satisfy $0<|h|<\delta$ and $x_{0}+h\in I$. By Step 2 and claim 10 of Elementary Order Arithmetic in an Ordered Field,

$$|b_{h}-a|<|h|\,\kappa<\delta_{2}\kappa=\mu ,$$

so $|b_{h}-a|<\eta$ and $|b_{h}-a|<A\cdot 2^{-1}$. By claims 3 and 7 of Properties of the Absolute Value in an Ordered Field,

$$A-|b_{h}|\le\bigl|\,|a|-|b_{h}|\,\bigr|\le|a-b_{h}|=|b_{h}-a|<A\cdot 2^{-1},$$

and since $A\cdot 2^{-1}+A\cdot 2^{-1}=A$ (claim 8 of Elementary Order Arithmetic in an Ordered Field), claim 1 there gives $A\cdot 2^{-1}<|b_{h}|$. Multiplying this strict inequality by the positive element $|b_{h}|^{-1}\bigl(2A^{-1}\bigr)$ (claims 5, 7 and 10 of Elementary Order Arithmetic in an Ordered Field) yields

$$|b_{h}|^{-1}<2A^{-1}.$$

By the choice of $\delta_{3}$ we also have $|Q(h)-L|<\eta$, and $A^{-1}|b_{h}-a|\,|L|\le A^{-1}\eta\,|L|$ by claim 5 of Elementary Arithmetic in an Ordered Field. Adding these (claim 3 of Elementary Order Arithmetic in an Ordered Field),

$$\bigl|Q(h)-L\bigr|+A^{-1}\bigl|b_{h}-a\bigr|\,|L|<\eta+A^{-1}\eta|L|=\eta\bigl(1+A^{-1}|L|\bigr).$$

Write $X=|b_{h}|^{-1}$, $X'=2A^{-1}$, $Y=|Q(h)-L|+A^{-1}|b_{h}-a||L|$ and $Y'=\eta(1+A^{-1}|L|)$; then $0\le X$, $0\le Y$, $X<X'$, $Y<Y'$ and $0<X'$, $0<Y'$. If $0<Y$ then $XY<X'Y<X'Y'$ by claim 10 of Elementary Order Arithmetic in an Ordered Field; if $Y=0$ then $XY=0<X'Y'$ by claim 5 there. In either case $XY<X'Y'$, and multiplying by the positive element $A^{-1}$ gives $A^{-1}XY<A^{-1}X'Y'$. Combining with the basic estimate of Step 1 and claim 2 of Elementary Order Arithmetic in an Ordered Field,

$$\bigl|R(h)+L(a^{-1})^{2}\bigr|\le A^{-1}XY<A^{-1}\bigl(2A^{-1}\bigr)\eta\bigl(1+A^{-1}|L|\bigr)=\eta C=\varepsilon .$$

Since $\varepsilon>0$ was arbitrary, Derivative at an Interior Point shows that $1/g$ is differentiable at $x_{0}$ with derivative $-L\bigl(a^{-1}\bigr)^{2}=-g'(x_{0})\bigl(g(x_{0})^{-1}\bigr)^{2}$, as claimed.

Proof of claim 2. Let $e:I\to\mathbb{R}$ be given by $e(z)=z$. By claim 1 of Properties of Natural Number Powers in a Field we have $z^{1}=z$ for the natural number powers of $\mathbb{R}$, so $e$ is the restriction to $I$ of the map $x\mapsto x^{1}$; by claim 1 of Derivative of a Polynomial Function on the Real Line, $e$ is differentiable at $x_{0}$ with $e'(x_{0})=1$. Since $0\notin I$ we have $e(z)=z\ne 0$ for every $z\in I$, and the function $1/e$ of claim 1 is exactly $r$. Claim 1 therefore applies and gives that $r$ is differentiable at $x_{0}$ with

$$r'(x_{0})=-1\cdot\bigl(x_{0}^{-1}\bigr)^{2}=-\bigl(x_{0}^{-1}\bigr)^{2}. \qquad\blacksquare$$