Proof of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers

Preliminaries. Continuity at $x$ relative to $I$ is read as in The Real Line: Standing Notation and Background for Calculus on Intervals (vi). The maps $\chi+\zeta$, $\chi\zeta$ and $a\chi$ of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, for $\chi,\zeta:I\to\mathbb{R}$ and $a\in\mathbb{R}$, are the pointwise operations of Pointwise Operations on Real-Valued Maps (i) (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)).

Two facts are used throughout. First, (D): if a map $\chi:I\to\mathbb{R}$ has a derivative $L$ at $x$, then $\chi$ is differentiable at $x$ and $\chi'(x)=L$; indeed $\chi$ is differentiable at $x$ by Derivative of a Real Function on an Interval (ii), and $\chi'(x)$ and $L$ are both derivatives of $\chi$ at $x$, so they are equal by Uniqueness of the Derivative of a Real Function on an Interval (i). Second, by (D) and The Increment Form of the Derivative, and Differentiability Implies Continuity (i), to show that $\chi$ is differentiable at $x$ with $\chi'(x)=L$ it suffices to give a map $\theta:I\to\mathbb{R}$, continuous at $x$ relative to $I$, with $\theta(x)=L$ and $\chi(t)-\chi(x)=\theta(t)\,(t-x)$ for every $t\in I$; such a $\theta$ is called an increment map for $\chi$ and $L$.

For $b\in\mathbb{R}$, write $k_{b}:I\to\mathbb{R}$ for the constant map with value $b$; it is continuous at $x$ relative to $I$ by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (i).

Proof of (i). For $t\in I$, $k_{b}(t)-k_{b}(x)=b-b=0=k_{0}(t)\,(t-x)$ (Difference and Quotient Identities in ℝ (i)), and $k_{0}(x)=0$. So $k_{0}$ is an increment map for $k_{b}$ and $0$.

Proof of (ii). Let $\iota:I\to\mathbb{R}$ be the map $t\mapsto t$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v)). For $t\in I$, $\iota(t)-\iota(x)=t-x=k_{1}(t)\,(t-x)$, and $k_{1}(x)=1$. So $k_{1}$ is an increment map for $\iota$ and $1$.

Proof of (iii). For $n\in\mathbb{N}_{0}$ let $p_{n}:I\to\mathbb{R}$ be the map $t\mapsto t^{n+1}$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v)), and let $A$ be the set of those $n\in\mathbb{N}_{0}$ for which $p_{n}$ is differentiable at $x$ with $p_{n}'(x)=(n+1)\,x^{n}$ (Sets and Maps: Ordinary Notation (iii)). Here and below, a natural number in a real expression is read in $\mathbb{R}$, and sums of natural numbers formed in $\mathbb{N}_{0}$ agree with those formed in $\mathbb{R}$, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vi). For $u\in\mathbb{R}$ and $m\in\mathbb{N}_{0}$ we have $u^{m+1}=u^{m}\,u^{1}=u^{m}\,u$, by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i) and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii), applied in the commutative ring $\mathbb{R}$.

$0\in A$: since $0+1=1$ in $\mathbb{N}_{0}$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (i), $p_{0}(t)=t^{1}=t$ for $t\in I$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii), so $p_{0}=\iota$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)). By (ii), $p_{0}$ is differentiable at $x$ with $p_{0}'(x)=1=1\cdot1=(0+1)\,x^{0}$, as $x^{0}=1$ by Powers with Exponents in the Natural Numbers with Zero (ii).

$n\in A$ implies $n+1\in A$: let $n\in A$. For $t\in I$, $p_{n+1}(t)=t^{(n+1)+1}=t^{n+1}\,t=p_{n}(t)\,\iota(t)$, so $p_{n+1}=p_{n}\iota$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)). The maps $p_{n}$ and $\iota$ are differentiable at $x$, by $n\in A$ and (ii). So (vi), applied to $p_{n}$ and $\iota$ in the roles of $f$ and $g$ (its proof below holds for any two maps $I\to\mathbb{R}$ differentiable at $x$ and does not use (iii)), shows that $p_{n+1}$ is differentiable at $x$ with

$$p_{n+1}'(x)=(n+1)\,x^{n}\,x+x^{n+1}\cdot1=(n+1)\,x^{n+1}+x^{n+1}=\bigl((n+1)+1\bigr)\,x^{n+1}.$$

So $n+1\in A$.

By induction from $0$ on $\mathbb{N}_{0}$, The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (viii), in which the successor of $n$ is $n+1$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (v), $A=\mathbb{N}_{0}$, which is (iii).

The maps $f$ and $g$. Let now $c\in\mathbb{R}$, and let $f,g:I\to\mathbb{R}$ be differentiable at $x$. Since $f'(x)$ and $g'(x)$ are derivatives of $f$ and $g$ at $x$ (Derivative of a Real Function on an Interval (ii)), The Increment Form of the Derivative, and Differentiability Implies Continuity (i) gives an increment map $\varphi$ for $f$ and $f'(x)$ and an increment map $\psi$ for $g$ and $g'(x)$. By The Increment Form of the Derivative, and Differentiability Implies Continuity (ii), $f$ and $g$ are continuous at $x$ relative to $I$.

Proof of (iv). For $t\in I$,

$$(f+g)(t)-(f+g)(x)=\bigl(f(t)-f(x)\bigr)+\bigl(g(t)-g(x)\bigr)=\bigl(\varphi(t)+\psi(t)\bigr)(t-x).$$

The map $\varphi+\psi$ is continuous at $x$ relative to $I$ by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (ii), and $(\varphi+\psi)(x)=f'(x)+g'(x)$. So it is an increment map for $f+g$ and $f'(x)+g'(x)$.

Proof of (v). For $t\in I$, $(cf)(t)-(cf)(x)=c\,\bigl(f(t)-f(x)\bigr)=\bigl(c\,\varphi(t)\bigr)(t-x)$. The map $c\varphi$ is continuous at $x$ relative to $I$ by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iv), and $(c\varphi)(x)=c\,f'(x)$. So it is an increment map for $cf$ and $c\,f'(x)$.

Proof of (vi). For $t\in I$,

$$(fg)(t)-(fg)(x)=\bigl(f(t)-f(x)\bigr)g(t)+f(x)\bigl(g(t)-g(x)\bigr)=\bigl(\varphi(t)\,g(t)+f(x)\,\psi(t)\bigr)(t-x).$$

Let $\theta=\varphi g+f(x)\psi$. The map $\varphi g$ is continuous at $x$ relative to $I$ by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iii), as $\varphi$ and $g$ are; $f(x)\psi$ is, by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iv); hence so is $\theta$, by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (ii). Moreover $\theta(x)=f'(x)\,g(x)+f(x)\,g'(x)$. So $\theta$ is an increment map for $fg$ and $f'(x)\,g(x)+f(x)\,g'(x)$.

Proof of (vii). Let $g(t)\neq0$ for every $t\in I$, and put $r=1/g$. For $t\in I$, $r(t)=1/g(t)=g(t)^{-1}$ and $g(t)\,r(t)=1$ (Difference and Quotient Identities in ℝ (ii)), and

$$(\dagger)\qquad r(t)-r(x)=\bigl(g(x)-g(t)\bigr)\,r(t)\,r(x),$$

since the right side is $g(x)\,r(x)\,r(t)-g(t)\,r(t)\,r(x)=r(t)-r(x)$, by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (ii) for $(u-v)\,w=u\,w-v\,w$ and by $g(x)\,r(x)=1=g(t)\,r(t)$.

Step 1: $r$ is continuous at $x$ relative to $I$. As $g$ is continuous at $x$ relative to $I$ and has no zeros, Continuity of Identity Maps, Compositions and Reciprocals (iii), applied with $E=I$, shows that the map $I\to\mathbb{R}$, $t\mapsto1/g(t)$, is continuous at $x$ relative to $I$; this map is $r$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)).

Step 2: the increment map. For $t\in I$, by $(\dagger)$, $g(x)-g(t)=-(g(t)-g(x))$ (Difference and Quotient Identities in ℝ (i)) and $g(t)-g(x)=\psi(t)\,(t-x)$,

$$r(t)-r(x)=-\psi(t)\,(t-x)\,r(t)\,r(x)=\bigl(-r(x)\,\psi(t)\,r(t)\bigr)(t-x).$$

Let $\theta=(-r(x))(\psi r)$. It is continuous at $x$ relative to $I$ by Step 1, Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iii) and Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iv). Moreover $\theta(x)=-g'(x)\,g(x)^{-1}\,g(x)^{-1}=-g'(x)/g(x)^{2}$ (Difference and Quotient Identities in ℝ (ii)), because $g(x)^{-1}\,g(x)^{-1}=(g(x)\,g(x))^{-1}=(g(x)^{2})^{-1}$ by Rules of Arithmetic and Order in an Ordered Field (iii), and $-(u\,v)=(-u)\,v$ by Rules of Arithmetic and Order in an Ordered Field (ii). So $\theta$ is an increment map for $r$ and $-g'(x)/g(x)^{2}$.

Proof of (viii). Let $g(t)\neq0$ for every $t\in I$, and $r=1/g$ as above. For $t\in I$, $(f/g)(t)=f(t)\,g(t)^{-1}=f(t)\,r(t)$ (Difference and Quotient Identities in ℝ (ii)), so $f/g=fr$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)). By (vii), $r$ is differentiable at $x$ with $r'(x)=-g'(x)/g(x)^{2}$. So (vi), applied to $f$ and $r$, shows that $f/g$ is differentiable at $x$ with

$$(f/g)'(x)=f'(x)\,g(x)^{-1}+f(x)\,\bigl(-g'(x)\bigr)\,(g(x)^{2})^{-1}=\bigl(f'(x)\,g(x)-f(x)\,g'(x)\bigr)/g(x)^{2},$$

since $g(x)^{-1}=g(x)\,(g(x)^{2})^{-1}$, which follows from $(g(x)^{2})^{-1}=g(x)^{-1}\,g(x)^{-1}$ (Rules of Arithmetic and Order in an Ordered Field (iii)) and $g(x)\,g(x)^{-1}=1$, and since $u\,(-v)=-(u\,v)$ by Rules of Arithmetic and Order in an Ordered Field (ii).