What is differentiable?

Example 15.4 (Radial functions, once and for all)

Let $r(x) = \norm x_2$ on $\R^n\setminus\{0\}$ and $f = g \circ r$ with $g$ a $C^1$ function of one variable. First, $r$ is differentiable away from $0$: from $r^2 = \sum x_i^2$,

the unit radial vector (differentiate $r^2$ and divide — or apply the chain rule to $\sqrt{\cdot}$). Then the chain rule gives, for every radial function,

Worked instance: $g(r) = \frac1r$ yields $\nabla\frac{1}{\norm x} = -\frac{x}{\norm x^3}$, the inverse-square field of gravitation and electrostatics — direction radial, magnitude $\frac{1}{\norm x^2}$. Closing insight: gradients of radial functions are radial because the level sets are spheres and the gradient is orthogonal to level sets; at $x = 0$, by contrast, $r$ is not differentiable (no candidate linear map matches $\norm h$ from all directions) — smooth radial profiles need $g'(0) = 0$ to cross the origin gracefully.

Example 15.6 (A Lipschitz constant from the MVI)

Is $f(x, y) = \sin x\,\sin y$ Lipschitz on $\R^2$, and with which constant? Its gradient is $\nabla f = (\cos x\sin y,\ \sin x\cos y)$, of squared norm

(bound $\cos^2x$ and $\sin^2x$ by $1$ separately), so $\vertiii{\dd f_{(x,y)}} = \norm{\nabla f} \leq 1$ everywhere, and Theorem 15.5 on the segment between any two points gives

$f$ is $1$-Lipschitz, and the constant is sharp (near the origin, $f(x, \tfrac\pi2) = \sin x$ has slope $1$). Closing insight: the mean value inequality converts a pointwise bound on the differential into a global modulus of continuity — the standard route to Lipschitz estimates in every dimension, and the engine inside Exercise 15.12.