Proof of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$, and $h$ always denotes a real number with $x_0+h\in I$. Recall from Derivative at an Interior Point that a real number $L$ is the derivative of a function $u:I\to\mathbb{R}$ at $x_0$ precisely when for every real $\varepsilon>0$ there is a real $\delta>0$ such that every such $h$ with $0<|h|<\delta$ satisfies $\bigl|\frac{u(x_0+h)-u(x_0)}{h}-L\bigr|<\varepsilon$.

Claim 1. Let $u$ be the function with constant value $b$. For every $h$ as above with $h\ne0$, $\frac{u(x_0+h)-u(x_0)}{h}=\frac{b-b}{h}=0$, so $\bigl|\frac{u(x_0+h)-u(x_0)}{h}-0\bigr|=0<\varepsilon$ for every real $\varepsilon>0$, with $\delta=1$ say. Hence $u$ is differentiable at $x_0$ with derivative $0$.

Claim 2. Write $A=f'(x_0)$ and $B=g'(x_0)$.

Sum. Let $\varepsilon>0$ be real. Choose $\delta_f>0$ for $f$ and $\varepsilon/2$, and $\delta_g>0$ for $g$ and $\varepsilon/2$, and set $\delta=\min\{\delta_f,\delta_g\}>0$. For $0<|h|<\delta$,

$$\frac{(f+g)(x_0+h)-(f+g)(x_0)}{h}-(A+B)=\Bigl(\frac{f(x_0+h)-f(x_0)}{h}-A\Bigr)+\Bigl(\frac{g(x_0+h)-g(x_0)}{h}-B\Bigr),$$

so by the triangle inequality the left side has absolute value less than $\varepsilon/2+\varepsilon/2=\varepsilon$.

Constant multiple. If $c=0$ then $cf$ is the constant function $0$, which by claim 1 is differentiable at $x_0$ with derivative $0=c\,A$. If $c\ne0$, let $\varepsilon>0$ be real and choose $\delta>0$ for $f$ and $\varepsilon/|c|$, which is positive. For $0<|h|<\delta$,

$$\Bigl|\frac{(cf)(x_0+h)-(cf)(x_0)}{h}-c\,A\Bigr|=|c|\,\Bigl|\frac{f(x_0+h)-f(x_0)}{h}-A\Bigr|<|c|\,\frac{\varepsilon}{|c|}=\varepsilon .$$

Claim 3. Again write $A=f'(x_0)$, $B=g'(x_0)$. For $h\ne0$ one has the algebraic identity

$$\frac{(fg)(x_0+h)-(fg)(x_0)}{h}-\bigl(A\,g(x_0)+f(x_0)\,B\bigr)=T_1+T_2+T_3,$$

where

$$T_1=f(x_0+h)\Bigl(\frac{g(x_0+h)-g(x_0)}{h}-B\Bigr),\quad T_2=B\bigl(f(x_0+h)-f(x_0)\bigr),\quad T_3=g(x_0)\Bigl(\frac{f(x_0+h)-f(x_0)}{h}-A\Bigr),$$

as one checks by expanding: $f(x_0+h)\frac{g(x_0+h)-g(x_0)}{h}+g(x_0)\frac{f(x_0+h)-f(x_0)}{h}$ equals the left-hand difference quotient, and the remaining terms cancel.

Since $f$ is differentiable at $x_0$, it is continuous at $x_0$ relative to $I$ by Differentiability at an Interior Point Implies Continuity There, the interval $I$ being regarded as a subset of the real line.

Let $\varepsilon>0$ be real. Set $M=|f(x_0)|+1$, so $M\ge1>0$. By the continuity just noted, applied with the positive real $\min\{1,\ \varepsilon/(3(|B|+1))\}$, there is $\delta_a>0$ such that every $h$ with $|h|<\delta_a$ satisfies

$$\bigl|f(x_0+h)-f(x_0)\bigr|<\min\Bigl\{1,\ \frac{\varepsilon}{3(|B|+1)}\Bigr\},$$

and hence also $|f(x_0+h)|\le|f(x_0)|+|f(x_0+h)-f(x_0)|<M$. Choose $\delta_b>0$ for $g$ and the positive real $\varepsilon/(3M)$, and $\delta_c>0$ for $f$ and the positive real $\varepsilon/(3(|g(x_0)|+1))$. Put $\delta=\min\{\delta_a,\delta_b,\delta_c\}>0$ and let $0<|h|<\delta$. Then

$$|T_1|<M\cdot\frac{\varepsilon}{3M}=\frac{\varepsilon}{3},\qquad |T_2|\le|B|\cdot\frac{\varepsilon}{3(|B|+1)}<\frac{\varepsilon}{3},\qquad |T_3|\le|g(x_0)|\cdot\frac{\varepsilon}{3(|g(x_0)|+1)}<\frac{\varepsilon}{3},$$

so by the triangle inequality the displayed difference has absolute value less than $\varepsilon$. As $\varepsilon>0$ was arbitrary, $fg$ is differentiable at $x_0$ with derivative $A\,g(x_0)+f(x_0)\,B$. $\blacksquare$