$r^2\sin (1\mathrm{/}r)$ is differentiable at the origin with a discontinuous gradient

Given: The function $F$ in the Example and $r=\parallel (x,y){\parallel }_2$.

For $(x,y)\in {\mathbb{R}}^2$, $\parallel (x,y){\parallel }_2=\sqrt{x^2+y^2}$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$).

For every real $t$, $\mathrm{\mid }\sin t\mathrm{\mid }\le 1$ and ${\sin }^{2}t+{\cos }^{2}t=1$ (Parity and the Pythagorean identity for sine and cosine).

The functions $\sin$ and $\cos$ are differentiable on $\mathbb{R}$, with $(\sin t{)}^{\mathrm{\prime }}=\cos t$ and $(\cos t{)}^{\mathrm{\prime }}=-\sin t$; also $\sin 0=0$ and $\cos 0=1$ (The derivatives of sine and cosine are cosine and minus sine).

If real functions $g$ and $h$ are differentiable at the relevant points, then $(g\circ h{)}^{\mathrm{\prime }}=(g^{\mathrm{\prime }}\circ h)h^{\mathrm{\prime }}$ (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 $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$).

Products, sums, scalar multiples, and quotients with nonzero denominator obey their usual derivative rules (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 $g(c)\mathrm{\ne }0$).

A map is totally differentiable at the origin with derivative zero when $\mathrm{\mid }F(h)\mathrm{\mid }\mathrm{/}\parallel h{\parallel }_2\to 0$ as $h\to 0$ through nonzero vectors (The total (Fréchet) derivative $Df(a)$ as the linear first-order approximation with $o(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$ remainder).

For a scalar function, the gradient is the vector of its coordinate partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

On $(0,\mathrm{\infty })$, $(u^{1\mathrm{/}2}{)}^{\mathrm{\prime }}=\frac{1}{2}{u}^{-1\mathrm{/}2}$ (Continuity and derivatives of positive-base real powers).

If $a>0$ and $q\in \mathbb{Q}$, then the real power $a^q$ agrees with the rational power; in particular this holds for $q=1\mathrm{/}2$ (The exponential definition of real powers agrees with the existing rational powers).

Every nonnegative real has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$).

For every real $t$, $\sin (t+\pi )=-\sin t$ and $\cos (t+\pi )=-\cos t$ (Quarter-turn values and shifts by pi/2 and pi).

Both sine and cosine have period $2\pi$ (The zero sets of sine and cosine and the least positive common period 2 pi).

For every real $\epsilon >0$ there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$ (For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$).

The number $\pi$ is positive (Pi as twice the smallest positive zero of cosine).

If a map is totally differentiable at a point, then each partial derivative there equals the total derivative applied to the corresponding standard basis vector (A total derivative computes every directional derivative, and its matrix is the Jacobian).