Rules for Derivatives of Vector- and Matrix-Valued Maps

In the settings of Single-Variable Calculus on an Interval and Euclidean Space and Real Matrices: Standing Notation, with $I$ as in Single-Variable Calculus on an Interval (ii), let $t\in I$ and $p,q,r,m,n\in\mathbb{N}$. Differentiability at $t$ and the derivative at $t$ are as in Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (i) and Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (ii) for maps from $I$ into $\mathbb{R}^{q}$ or $\mathcal{M}_{p\times q}(\mathbb{R})$, and as in Derivative of a Real Function on an Interval (ii) for maps $I\to\mathbb{R}$; continuity at $t$ is continuity at $t$ relative to $I$, with $I\subseteq\mathbb{R}$ carrying $d_{\mathbb{R}}$ as in Single-Variable Calculus on an Interval (iii), $d_{E}$ on $\mathbb{R}^{q}$ and $d_{p\times q}$ on $\mathcal{M}_{p\times q}(\mathbb{R})$. Sums, differences, negatives, products with real maps, dot products, matrix products, matrix-vector products and transposes of maps on $I$, such as $x+y$, $x-y$, $-x$, $\varphi x$, $x\cdot y$, $AB$, $Ax$ and $A^{\top}$, are formed pointwise, e.g. $(x+y)(s)=x(s)+y(s)$.

(i) If $x:I\to\mathbb{R}^{q}$ is differentiable at $t$, then $x$ is continuous at $t$; if $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $t$, then $A$ is continuous at $t$.

(ii) If $x,y:I\to\mathbb{R}^{q}$ are differentiable at $t$, then so is $x+y$, and $(x+y)'(t)=x'(t)+y'(t)$. If $A,B:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ are differentiable at $t$, then so is $A+B$, and $(A+B)'(t)=A'(t)+B'(t)$.

(iii) Let $\varphi:I\to\mathbb{R}$ be differentiable at $t$. If $x:I\to\mathbb{R}^{q}$ is differentiable at $t$, then so is $\varphi x$, and $(\varphi x)'(t)=\varphi'(t)\,x(t)+\varphi(t)\,x'(t)$. If $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $t$, then so is $\varphi A$, and $(\varphi A)'(t)=\varphi'(t)\,A(t)+\varphi(t)\,A'(t)$.

(iv) If $x:I\to\mathbb{R}^{q}$ is differentiable at $t$, then so is $-x$, and $(-x)'(t)=-x'(t)$; if moreover $y:I\to\mathbb{R}^{q}$ is differentiable at $t$, then so is $x-y$, and $(x-y)'(t)=x'(t)-y'(t)$. If $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $t$, then so is $-A$, and $(-A)'(t)=-A'(t)$; if moreover $B:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $t$, then so is $A-B$, and $(A-B)'(t)=A'(t)-B'(t)$.

(v) If $A:I\to\mathcal{M}_{p\times r}(\mathbb{R})$ and $B:I\to\mathcal{M}_{r\times q}(\mathbb{R})$ are differentiable at $t$, then so is $AB$, and $(AB)'(t)=A'(t)B(t)+A(t)B'(t)$.

(vi) If $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ and $x:I\to\mathbb{R}^{q}$ are differentiable at $t$, then so is $Ax$, and $(Ax)'(t)=A'(t)x(t)+A(t)x'(t)$.

(vii) If $x,y:I\to\mathbb{R}^{q}$ are differentiable at $t$, then so is $x\cdot y:I\to\mathbb{R}$, and $(x\cdot y)'(t)=x'(t)\cdot y(t)+x(t)\cdot y'(t)$.

(viii) If $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $t$, then so is $A^{\top}:I\to\mathcal{M}_{q\times p}(\mathbb{R})$, and $(A^{\top})'(t)=A'(t)^{\top}$.

(ix) For $b\in\mathbb{R}^{q}$ and $C\in\mathcal{M}_{p\times q}(\mathbb{R})$, the constant maps $I\to\mathbb{R}^{q}$, $s\mapsto b$, and $I\to\mathcal{M}_{p\times q}(\mathbb{R})$, $s\mapsto C$, are differentiable at $t$ with derivatives $0$ and $0_{p\times q}$.

(x) Let $L\in\mathcal{M}_{m\times q}(\mathbb{R})$, $z\in\mathbb{R}^{q}$, $M\in\mathcal{M}_{m\times p}(\mathbb{R})$ and $N\in\mathcal{M}_{q\times n}(\mathbb{R})$. If $x:I\to\mathbb{R}^{q}$ is differentiable at $t$, then the maps $s\mapsto Lx(s)$ and $s\mapsto z\cdot x(s)$ are differentiable at $t$, with derivatives $Lx'(t)$ and $z\cdot x'(t)$. If $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $t$, then the maps $s\mapsto A(s)z$ and $s\mapsto\bigl(MA(s)\bigr)N$ are differentiable at $t$, with derivatives $A'(t)z$ and $\bigl(MA'(t)\bigr)N$.

(xi) Let $J$ be an interval containing two distinct points, let $\psi:J\to\mathbb{R}$ satisfy $\psi(\sigma)\in I$ for every $\sigma\in J$, and let $s_{0}\in J$ be such that $\psi$ is differentiable at $s_{0}$ (relative to $J$). If $x:I\to\mathbb{R}^{q}$ is differentiable at $\psi(s_{0})$, then $x\circ\psi:J\to\mathbb{R}^{q}$ is differentiable at $s_{0}$ (relative to $J$), and $(x\circ\psi)'(s_{0})=\psi'(s_{0})\,x'(\psi(s_{0}))$. If $A:I\to\mathcal{M}_{p\times q}(\mathbb{R})$ is differentiable at $\psi(s_{0})$, then $A\circ\psi$ is differentiable at $s_{0}$, and $(A\circ\psi)'(s_{0})=\psi'(s_{0})\,A'(\psi(s_{0}))$.