# Abstract Nonsense

## Further Properties of the Total Derivative (Pt. II)

Point of Post: This is a continuation of this post.

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We now prove a similar theorem for products of real-valued functions

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Theorem: Let $f_1,\cdots,f_k:U\to\mathbb{R}$ with $U\subseteq\mathbb{R}^n$ be differentiable at $a\in U$. Then, $f_1\cdots f_k:U\to\mathbb{R}$ given by $(f_1\cdots f_k)(x)=f_1(x)\cdots f_k(x)$ is differentiable at $a$ and

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$\displaystyle D_{f_1\cdots f_k}(a)=\sum_{j=1}^{k}D_{f_1}(a)\cdots D_{f_j}(a)\cdots D_{f_k}(a)$

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Proof: We play a similar game to the sum proof. Namely, consider $K:\mathbb{R}^n\to\mathbb{R}^k$ given by $x\mapsto (f_1(x),\cdots,f_k(x))$ and $M:\mathbb{R}^k\to\mathbb{R}$ given by $(x_1,\cdots,x_k)\mapsto x_1\cdots x_k$ and note that $f_1\cdots f_k=M\circ K$. Now, $K$ is differentiable at $a$ since each coordinate function is differentiable at $a$ and $M$ is differentiable at $a$ since it’s mutlilinear. Thus, by the chain rule we have that $f_1\cdots f_k$ is differentiable at $a$ and $D_{f_1\cdots f_k}(a)=D_{M}(K(a))\circ D_K(a)$. But, using the same logic as for the previous theorem we have that

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$D_K(a)=\left(\begin{array}{ccc} & \text{Jac}_{f_1}(a)\\ \hline & \vdots & \\ \hline & \text{Jac}_{f_k}(a)\end{array}\right)$

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And from our corollary concerning the total derivative of multilinear functions we know that

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$\displaystyle D_K(K(a))(x)=\sum_{j=1}^{k}f_1(a)\cdots x_j\cdots f_n(a)$

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where evidently $x=(x_1,\cdots,x_n)$. Thus, with this in mind we see that for every $x\in\mathbb{R}$ we have

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\displaystyle \begin{aligned}D_{f_1\cdots f_k}(a)(x) &=\left(D_M(K(a))\right)\left(D_K(a)(x)\right)\left(D_M(K(a))\right)\left(\left(\text{Jac}_{f_1}(a)(x),\cdots,\text{Jac}_{f_k}(a)(x)\right)\right)\\ &=\sum_{j=1}^{k}f_1(a)\cdots\text{Jac}_{f_j}(a)\cdots f_{k}(a)\\ &= \sum_{j=1}^{k}f_1(a)\cdots D_{f_j}(a)(x)\cdots f_k(a)\end{aligned}

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We now compute the derivative of the inner product of two differentiable functions. In particular:

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Theorem: Let $f,g:U\to\mathbb{R}^m$ where $U\subseteq\mathbb{R}^n$ be differentiable at $a\in U$. Then, the function $F:U\to\mathbb{R}$  given by $F:x\mapsto \left\langle f(x),g(x)\right\rangle$ (where $\left\langle\cdot,\cdot\right\rangle$ is the usual inner product (i.e. dot product) on $\mathbb{R}^m$) is differentiable at $a$ and $D_F(a)(x)=\left\langle D_f(a)(x),g(a)\right\rangle+\left\langle f(a),D_g(a)(x)\right\rangle$

Proof: We play the same old game. Namely define $K:\mathbb{R}^n\to\mathbb{R}^{2m}$ given by $x\mapsto (f_1(x),\cdots,f_m(x),g_1(x),\cdots,g_m(x))$ where $f_j,g_j$ are the coordinate functions of $f,g$ respectively. We know since each coordinate function is differentiable at $a$ so is $K$ and moreover

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$D_K(a)=\left(\begin{array}{ccc} & \text{Jac}_{f_1}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_m}(a)\\ \hline & \text{Jac}_{g_1}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{g_m}(a) & \end{array}\right)$

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We also know from a past corollary that $\langle\cdot,\cdot\rangle:\mathbb{R}^{2m}\to\mathbb{R}$ is differentiable everywhere and

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$\displaystyle D_{\langle\cdot,\cdot\rangle}((x,y))(z)=\left\langle x,z\right\rangle+\left\langle z,y\right\rangle$

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(where of course we use shorthand notation meaning $x,y\in\mathbb{R}^m$). Noticing then that $\langle\cdot\cdot\rangle\circ K=F$ we may conclude from the chain rule that $F$ is differentiable at $a$ and $D_F(a)=D_{\langle\cdot,\cdot\rangle}(K(a))\circ D_K(a)$. Thus,

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\begin{aligned}D_F(a)(x)&=\left(D_{\langle\cdot,\cdot\rangle}(K(a))\right)\left(D_K(a)\right)\\ &=\left(D_{\langle\cdot\cdot\rangle}(K(a))\right)\left(\text{Jac}_{f_1}(a)(x),\cdots,\text{Jac}_{f_m}(a)(x),\text{Jac}_{g_1}(a)(x),\cdots,\text{Jac}_{g_m}(a)(x)\right)\\ &=\left\langle \left(\text{Jac}_{f_1}(a)(x),\cdots,\text{Jac}_{f_m}(a)(x)\right),\left(\text{Jac}_{g_1}(a)(x),\cdots,\text{Jac}_{g_m}(a)(x)\right)\right\rangle\\ &= \left\langle \text{Jac}_{f}(a)(x),g(a)\right\rangle+\left\langle f(a),\text{Jac}_{g}(a)(x)\right\rangle\\ &= \left\langle D_f(a)(x),g(a)\right\rangle+\left\langle f(a),D_g(a)(x)\right\rangle\end{aligned}

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since $x$ was arbitrary the conclusion follows. $\blacksquare$

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References:

1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.