Sum, product, quotient, and chain rules from difference quotients

If $f$ is differentiable at $c$, write $f(c+h)-f(c)=hq(h)$, where $q(h)\to f^{\mathrm{\prime }}(c)$. Then $q$ is bounded for sufficiently small $h$, so $\mathrm{\mid }f(c+h)-f(c)\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }q(h)\mathrm{\mid }\to 0$. Thus $f$ is continuous at $c$.

The sum difference quotient is the sum of the two difference quotients, so it tends to $f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$. For the product, use $\frac{f(c+h)g(c+h)-f(c)g(c)}{h}=f(c+h)\frac{g(c+h)-g(c)}{h}+g(c)\frac{f(c+h)-f(c)}{h}\mathrm{.}$ By step 1.1, $f(c+h)\to f(c)$; taking limits gives the product rule.

By step 1.1, $g$ is continuous at $c$. Since $g(c)\mathrm{\ne }0$, $g(c+h)\mathrm{\ne }0$ for sufficiently small $h$. Algebra gives $\frac{f(c+h)\mathrm{/}g(c+h)-f(c)\mathrm{/}g(c)}{h}=\frac{g(c)(f(c+h)-f(c))\mathrm{/}h-f(c)(g(c+h)-g(c))\mathrm{/}h}{g(c+h)g(c)}\mathrm{.}$ The denominator tends to $g(c{)}^2\mathrm{\ne }0$ and the numerator tends to $g(c)f^{\mathrm{\prime }}(c)-f(c)g^{\mathrm{\prime }}(c)$, proving the quotient rule.

Put $y=f(c)$ and $\mathrm{\Delta }(h)=f(c+h)-y$. Step 1.1 gives $\mathrm{\Delta }(h)\to 0$, and [F1] gives $\mathrm{\Delta }(h)\mathrm{/}h\to f^{\mathrm{\prime }}(c)$. Differentiability of $g$ at $y$ means $g(y+u)-g(y)=g^{\mathrm{\prime }}(y)u+uϵ(u)$ for $u\mathrm{\ne }0$, where $ϵ(u)\to 0$; set $ϵ(0)=0$. Then $\frac{g(f(c+h))-g(f(c))}{h}=g^{\mathrm{\prime }}(y)\frac{\mathrm{\Delta }(h)}{h}+ϵ(\mathrm{\Delta }(h))\frac{\mathrm{\Delta }(h)}{h}\mathrm{.}$ The first term tends to $g^{\mathrm{\prime }}(y)f^{\mathrm{\prime }}(c)$; the second tends to zero because $ϵ(\mathrm{\Delta }(h))\to 0$ and $\mathrm{\Delta }(h)\mathrm{/}h$ is bounded. This proves the chain rule, including increments with $\mathrm{\Delta }(h)=0$.

The case $m=1$ is the derivative of the identity function, whose quotient is exactly $1$. Suppose $(x^m{)}^{\mathrm{\prime }}=m{x}^{m-1}$. Since $x^{m+1}=x\cdot x^m$, the product rule in step 2.1 gives $(x^{m+1}{)}^{\mathrm{\prime }}=x^m+x(m{x}^{m-1})=(m+1)x^m$. Induction proves the power rule for all $m\ge 1$.