Gradient of the distance is the outward unit radial field off the base point and the cut locus

Setup, and the identification of $u$ with $\dot{\gamma }(t)$. [A1, F1, F2, F4, F5, F6, F7, F11, F24, F25] By [F4] and [F5], ${\mathcal{E}}_p=T_{p}M$ and ${\exp }_p:T_{p}M\to M$ is smooth. By [F6], $t\in I_{p,v}$ and ${\exp }_p(tv)=\gamma (t)$; by [F7], $\gamma$ is the unique maximal geodesic with $\gamma (0)=p$ and $\dot{\gamma }(0)=v$, and it is smooth. Since $v\in S_{p}M$ and $0<t<c_p(v)$, the vector $w_0=tv$ lies in $D_p$, so [F2] puts $q={\exp }_p(w_0)$ in the open submanifold $U:=M\setminus (\{p\}\cup \mathrm{Cut}(p))$, on which $r_p$ is smooth by [F1]; hence $\mathrm{grad}{r}_p(q)$ is defined by [F19]. The segment $\gamma {\mathrm{\mid }}_{[0,t]}$ is minimizing by [F24]. For the velocity, consider the smooth curve $c(s):=w_0+sv$ in the vector space $T_{p}M$, so that $c(0)=w_0$ and $\dot{c}(0)=v$ under the canonical identification $T_{w_0}(T_{p}M)\cong T_{p}M$ [F25]. Since ${\exp }_p(c(s))={\exp }_p((t+s)v)=\gamma (t+s)$ by [F6], [F11] gives $d({\exp }_p{)}_{w_0}(v)=(d\mathrm{/}ds){\mathrm{\mid }}_0({\exp }_p\circ c)(s)=\dot{\gamma }(t)$, that is, $u=\dot{\gamma }(t)$ [F25].

The composite and its differential. [F1, F2, F5, F8, F9, F10, F25] By [F1], $r_p({\exp }_p(w))=\mathrm{\mid }w{\mathrm{\mid }}_g$ for every $w\in D_p$, and by [F2] the set $D_p$ is open with $w_0\in D_p$ and ${\exp }_p(w_0)=q$. Both ${\exp }_p$ (smooth on $T_{p}M$ by [F5] and [F4]) and $r_p$ (smooth on $U$ by [F1]) are smooth, so [F10] applies at $w_0$ and gives $d(r_p\circ {\exp }_p{)}_{w_0}=d(r_p{)}_q\circ d({\exp }_p{)}_{w_0}$. Because $r_p\circ {\exp }_p$ agrees on the open set $D_p$ with the function $N_p$ of [F9], which is smooth on $T_{p}M\setminus \{0_p\}$ by [F8] and defined at the nonzero vector $w_0$ (as $t>0$ and $v\mathrm{\ne }0$), we obtain the identity of linear maps $d(r_p{)}_q\circ d({\exp }_p{)}_{w_0}=d(N_p{)}_{w_0}:T_{p}M\to \mathbb{R},$ where both sides are read through the canonical identification $T_{w_0}(T_{p}M)\cong T_{p}M$ [F25].

The derivative of the squared norm along a line. [F14, F15, F18] Fix $X\in T_{p}M$ and put $Q(w):=g_p(w,w)$, so that $Q$ is a quadratic form on $T_{p}M$ [F18]. Along $c(s)=w_0+sX$, bilinearity and symmetry of $g_p$ [F18] give $Q(c(s))=g_p(w_0,w_0)+2s{g}_p(w_0,X)+s^2{g}_p(X,X);$ the right-hand side is a polynomial in $s$ with constant term $Q(w_0)$, linear coefficient $2g_p(w_0,X)$ and quadratic coefficient $Q(X)$, so by the sum, scalar and product rules [F14] together with the power derivatives $p_0^{\mathrm{\prime }}=0$, $p_1^{\mathrm{\prime }}=1$, $p_2^{\mathrm{\prime }}(0)=0$ [F15] its derivative at $0$ is $2g_p(w_0,X)$. Since $Q\circ c$ is that polynomial, $(Q\circ c{)}^{\mathrm{\prime }}(0)=2g_p(w_0,X)$; this is a statement about one real function of $s$ and needs no smoothness of $Q$ beyond the polynomial displayed.

The differential of the pointwise norm. [F8, F9, F11, F12, F13, F18, F20, F21, F22, F23, step 1.3] Fix $X\in T_{p}M$ and let $c(s)=w_0+sX$ as in step 1.3. Since $c(0)=w_0\mathrm{\ne }0$, the vector $c(s)$ is nonzero for all sufficiently small $s$; for those $s$ the pointwise norm is positive and, by [F9] and [F21], $N_p(c(s))=g_p(c(s),c(s){)}^{1\mathrm{/}2}=(Q(c(s)){)}^{1\mathrm{/}2}\mathrm{.}$ Write $\psi :=Q\circ c$, a real function differentiable at $0$ with $\psi (0)=Q(w_0)=\mathrm{\mid }{w}_0{\mathrm{\mid }}_g^2=(t\mathrm{\mid }v{\mathrm{\mid }}_g{)}^2=t^2>0$, the homogeneity $\mathrm{\mid }tv{\mathrm{\mid }}_g=t\mathrm{\mid }v{\mathrm{\mid }}_g$ being [F20], and with ${\psi }^{\mathrm{\prime }}(0)=2g_p(w_0,X)$ by step 1.3. The real chain rule [F13] applied to $N_p\circ c={\psi }^{1\mathrm{/}2}$ and the power derivative [F12] with exponent $\alpha =1\mathrm{/}2$ give $(N_p\circ c{)}^{\mathrm{\prime }}(0)=\frac{1}{2}\psi (0{)}^{-1\mathrm{/}2}{\psi }^{\mathrm{\prime }}(0)=\frac{g_p(w_0,X)}{\psi (0{)}^{1\mathrm{/}2}}\mathrm{.}$ Here $\psi (0{)}^{1\mathrm{/}2}=(t^2{)}^{1\mathrm{/}2}=t^{2\cdot (1\mathrm{/}2)}=t^1=t$ by [F22] and [F23], and $g_p(w_0,X)=t{g}_p(v,X)$ by bilinearity [F18] with $w_0=tv$, so $(N_p\circ c{)}^{\mathrm{\prime }}(0)=g_p(v,X)$. By [F11], $d(N_p{)}_{w_0}(X)=g_p(v,X)\mathrm{.}$

The pairing identity and the conclusion. [F3, F16, F17, F18, F19, step 1.2, step 2.1] Fix $X\in T_{p}M$ and put $Y:=d({\exp }_p{)}_{w_0}(X)\in T_{q}M$. Gauss's lemma [F3] applied at the base vector $w_0\in {\mathcal{E}}_p=T_{p}M$ with the vector $X$ gives $g_q(d({\exp }_p{)}_{w_0}(w_0),d({\exp }_p{)}_{w_0}(X))=g_p(w_0,X)$. Since $w_0=tv$, linearity of the differential [F17] gives $d({\exp }_p{)}_{w_0}(w_0)=td({\exp }_p{)}_{w_0}(v)=tu$, so bilinearity of $g_q$ [F18] yields $g_q(u,Y)=\frac{1}{t}{g}_q(d({\exp }_p{)}_{w_0}(w_0),Y)=\frac{g_p(w_0,X)}{t}=g_p(v,X)\mathrm{.}$ On the other hand steps 1.2 and 2.1 give $d(r_p{)}_q(Y)=d(N_p{)}_{w_0}(X)=g_p(v,X)$. Hence $g_q(u,Y)=d(r_p{)}_q(Y)$ for every $Y$ in the image of $d({\exp }_p{)}_{w_0}$, and that image is all of $T_{q}M$ because ${\exp }_p{\mathrm{\mid }}_{D_p}$ is a diffeomorphism onto its image [F2] and the differential of a diffeomorphism is a linear isomorphism [F16]; so $g_q(u,Y)=d(r_p{)}_q(Y)\text{for every }Y\in T_{q}M\mathrm{.}$ Taking $X:=v$ in the same computation gives $Y=u$ and $g_q(u,u)=g_p(v,v)=1$; hence $\mathrm{\mid }u{\mathrm{\mid }}_g=\sqrt{g_q(u,u)}=1$ by [F9] and [F21]. Finally, [F19] says that $(\mathrm{grad}{r}_p)(q)$ is the vector with $g_q((\mathrm{grad}{r}_p)(q),Y)=d(r_p{)}_q(Y)$ for every $Y$; if two vectors $u,u^{\mathrm{\prime }}$ have this property then $g_q(u-u^{\mathrm{\prime }},Y)=0$ for every $Y$ by bilinearity [F18], and $Y:=u-u^{\mathrm{\prime }}$ gives $g_q(u-u^{\mathrm{\prime }},u-u^{\mathrm{\prime }})=0$, so $u=u^{\mathrm{\prime }}$ by positive definiteness [F18]. Therefore $\mathrm{grad}{r}_p(q)=u=\dot{\gamma }(t)=d({\exp }_p{)}_{tv}(v)$ [step 1.1], a unit vector.

Boundary, endpoint and choice audit. [A1, F1, F2, F3, F4, F24, step 1.1, step 3.1] In dimension zero $S_{p}M=\mathrm{\varnothing }$, there is no unit direction $v$, and the assertion is vacuous; the empty manifold carries no base point $p$. The parameter range $0<t<c_p(v)$ is open, and both endpoints are genuinely excluded: at $t=0$ the point is $q=p$, where $r_p$ is not smooth and $\mathrm{grad}{r}_p$ is not defined, while at $t=c_p(v)<+\mathrm{\infty }$ the point $q$ lies in $\mathrm{Cut}(p)$ and $tv\mathrm{\notin }{D}_p$, so the restricted diffeomorphism ${\exp }_p{\mathrm{\mid }}_{D_p}$ of [F2] does not include $q$; the infinite cut time $c_p(v)=+\mathrm{\infty }$ is allowed, and then every $t>0$ is covered by [F24] and [F2]. The unit hypothesis $\mathrm{\mid }v{\mathrm{\mid }}_g=1$ is exactly what makes $u$ unit, while $t>0$ permits division by $t$. For any $W\in D_p$ (including nonunit $W$), set $v=W\mathrm{/}\mathrm{\mid }W{\mathrm{\mid }}_g$ and $t=\mathrm{\mid }W{\mathrm{\mid }}_g$; then $0<t<c_p(v)$ and the same computation gives $\mathrm{grad}{r}_p({\exp }_p(W))=d({\exp }_p{)}_W(W)\mathrm{/}\mathrm{\mid }W{\mathrm{\mid }}_g$, the radial unit field. Degenerate directions $X=0$ are harmless: then $Y=0$, both sides vanish, and step 1.3's polynomial has $B=C=0$. The zero vector is never used, because $w_0\mathrm{\ne }0$ and $N_p$ is smooth off zero [F3, F8]. Assumption [A1] is inherited exactly through the exponential, cut-time, Hopf-Rinow, Gauss and distance suppliers, and no further selection is made: $X$ is fixed, the inverse of the linear isomorphism in step 3.1 is a function, and no choice function is invoked. The statement is an equality of vectors with a norm assertion; it contains no biconditional, and the only equivalence invoked, [F6], is used in both directions of its stated formula for the single pair $(p,v)$. Compactness of $M$ and positivity of the injectivity radius are never used.