Proof of Sum and Product Rules for One-Dimensional Derivatives and Continuity

Throughout, differentiability at an interior point $x_0$ of $I$ is the condition of Derivative at an Interior Point; its admissible increments are the real numbers $h$ with $x_0+h\in I$. Continuity is the condition of Continuity at a Point. We repeatedly use the triangle inequality for real numbers.

Claim 1. Write $L=f'(x_0)$. Applying the definition of the derivative with $\varepsilon=1$, there is $\delta_0>0$ such that for every admissible $h$ with $0<|h|<\delta_0$,

$$\Bigl|\frac{f(x_0+h)-f(x_0)}{h}-L\Bigr|<1,$$

hence $|f(x_0+h)-f(x_0)|\le(|L|+1)|h|$. Let $\varepsilon>0$ and put $\delta=\min\bigl(\delta_0,\ \varepsilon/(|L|+1)\bigr)$. For $x\in I$ with $|x-x_0|<\delta$: if $x=x_0$ then $|f(x)-f(x_0)|=0<\varepsilon$; otherwise $x=x_0+h$ with $h=x-x_0$ admissible and $0<|h|<\delta$, so

$$|f(x)-f(x_0)|\le(|L|+1)|h|<\varepsilon.$$

Hence $f$ is continuous at $x_0$.

Claim 2. Let $\varepsilon>0$. Choose $\delta_1>0$ for $f$ and $\delta_2>0$ for $g$ so that the respective difference quotients at $x_0$ are within $\varepsilon/2$ of $f'(x_0)$ and of $g'(x_0)$ for admissible $h$ with $0<|h|<\delta_1$ (respectively $\delta_2$). The difference quotient of $f+g$ at $x_0$ is the sum of the two difference quotients, so for admissible $h$ with $0<|h|<\min(\delta_1,\delta_2)$,

$$\Bigl|\frac{(f+g)(x_0+h)-(f+g)(x_0)}{h}-\bigl(f'(x_0)+g'(x_0)\bigr)\Bigr|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

If $f$ is constant, every difference quotient equals $0$, so the defining condition holds with derivative $0$ for every $\delta$; in particular for $c=0$ the function $cf$ is the constant $0$ and $(cf)'(x_0)=0=c\,f'(x_0)$. For $c\ne0$, choose $\delta_3>0$ so that the difference quotient of $f$ is within $\varepsilon/|c|$ of $f'(x_0)$ for admissible $0<|h|<\delta_3$; multiplying by $c$ shows that the difference quotient of $cf$ is within $\varepsilon$ of $c\,f'(x_0)$.

Claim 3. By Claim 1 applied to $g$, with $\varepsilon=1$ in the continuity condition there is $\delta_g>0$ such that $|g(x_0+h)-g(x_0)|<1$, and hence $|g(x_0+h)|\le K$ with $K=|g(x_0)|+1\ge1$, for every admissible $h$ with $|h|<\delta_g$. For admissible $h\ne0$, the identity

$$f(x_0+h)g(x_0+h)-f(x_0)g(x_0)=\bigl(f(x_0+h)-f(x_0)\bigr)g(x_0+h)+f(x_0)\bigl(g(x_0+h)-g(x_0)\bigr)$$

gives, after dividing by $h$ and subtracting $f'(x_0)g(x_0)+f(x_0)g'(x_0)$,

$$\frac{(fg)(x_0+h)-(fg)(x_0)}{h}-f'(x_0)g(x_0)-f(x_0)g'(x_0) =\Bigl(\frac{f(x_0+h)-f(x_0)}{h}-f'(x_0)\Bigr)g(x_0+h)+f'(x_0)\bigl(g(x_0+h)-g(x_0)\bigr)+f(x_0)\Bigl(\frac{g(x_0+h)-g(x_0)}{h}-g'(x_0)\Bigr).$$

Let $\varepsilon>0$. Choose $\delta_a>0$ so that the difference quotient of $f$ is within $\varepsilon/(3K)$ of $f'(x_0)$ for admissible $0<|h|<\delta_a$; choose $\delta_b>0$ so that $|g(x_0+h)-g(x_0)|<\varepsilon/\bigl(3(|f'(x_0)|+1)\bigr)$ for admissible $|h|<\delta_b$ (possible by Claim 1 applied to $g$); choose $\delta_c>0$ so that the difference quotient of $g$ is within $\varepsilon/\bigl(3(|f(x_0)|+1)\bigr)$ of $g'(x_0)$ for admissible $0<|h|<\delta_c$. For admissible $h$ with $0<|h|<\min(\delta_g,\delta_a,\delta_b,\delta_c)$, the three terms on the right of the last display are bounded in absolute value by $\varepsilon/3$ each, so the difference quotient of $fg$ is within $\varepsilon$ of $f'(x_0)g(x_0)+f(x_0)g'(x_0)$. This proves the product rule.

Claim 4. Let $\varepsilon>0$. For $\varphi+\psi$: choose $\delta_1,\delta_2>0$ with $|\varphi(x)-\varphi(x_0)|<\varepsilon/2$ for all $x\in E$ with $|x-x_0|<\delta_1$, and $|\psi(x)-\psi(x_0)|<\varepsilon/2$ for all $x\in E$ with $|x-x_0|<\delta_2$; then for $x\in E$ with $|x-x_0|<\min(\delta_1,\delta_2)$,

$$|(\varphi+\psi)(x)-(\varphi+\psi)(x_0)|\le|\varphi(x)-\varphi(x_0)|+|\psi(x)-\psi(x_0)|<\varepsilon.$$

For $c\varphi$: choose $\delta>0$ with $|\varphi(x)-\varphi(x_0)|<\varepsilon/(|c|+1)$ for $x\in E$ with $|x-x_0|<\delta$; then $|c\varphi(x)-c\varphi(x_0)|\le|c|\,\varepsilon/(|c|+1)<\varepsilon$. For $\varphi\psi$: use the identity

$$\varphi(x)\psi(x)-\varphi(x_0)\psi(x_0)=\bigl(\varphi(x)-\varphi(x_0)\bigr)\psi(x)+\varphi(x_0)\bigl(\psi(x)-\psi(x_0)\bigr).$$

Choose $\delta_2>0$ so that $|\psi(x)-\psi(x_0)|<\min\Bigl(1,\ \frac{\varepsilon}{2(|\varphi(x_0)|+1)}\Bigr)$ for $x\in E$ with $|x-x_0|<\delta_2$, so that also $|\psi(x)|\le|\psi(x_0)|+1$ there, and choose $\delta_1>0$ so that $|\varphi(x)-\varphi(x_0)|<\frac{\varepsilon}{2(|\psi(x_0)|+1)}$ for $x\in E$ with $|x-x_0|<\delta_1$. For $x\in E$ with $|x-x_0|<\min(\delta_1,\delta_2)$, the first term is smaller than $\varepsilon/2$ in absolute value and the second is at most $|\varphi(x_0)|\cdot\frac{\varepsilon}{2(|\varphi(x_0)|+1)}<\varepsilon/2$, so $|\varphi(x)\psi(x)-\varphi(x_0)\psi(x_0)|<\varepsilon$. Finally, for a constant function the difference of values at $x$ and $x_0$ is $0$ for all $x\in E$, so any $\delta$ works. $\blacksquare$