$F(x)=x^2\sin (1\mathrm{/}{x}^2)$ has an unbounded derivative whose Henstock–Kurzweil integral is $\sin 1$

Given: The displayed function $F$ and its derivative candidate $f$.

If $a<b$, $F:[a,b]\to \mathbb{R}$ is differentiable in the domain-relative sense, including at the endpoints, and $f=F^{\mathrm{\prime }}$, then $f$ is Henstock–Kurzweil integrable and ${\int }_a^{b}f=F(b)-F(a)$ (Every derivative is Henstock–Kurzweil integrable and satisfies Newton–Leibniz).

For every integer $m$, $\sin (\pi \mathrm{/}2+2m\pi )=1$, $\sin (2m\pi )=0$, and $\cos (2m\pi )=1$ (Quarter-turn values and shifts by pi/2 and pi, The zero sets of sine and cosine and the least positive common period 2 pi).

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the chain rule gives $(f\circ g{)}^{\mathrm{\prime }}(x)=f^{\mathrm{\prime }}(g(x))g^{\mathrm{\prime }}(x)$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$).

The product rule holds for differentiable real functions, and the quotient rule holds where the denominator is nonzero (Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$).

Sine and cosine have derivatives $\cos$ and $-\sin$ (The derivatives of sine and cosine are cosine and minus sine).

For every real $u$, $\mathrm{\mid }\sin u\mathrm{\mid }\le 1$ (Parity and the Pythagorean identity for sine and cosine).