# 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 showed all of these proofs because it’s helpful to go through them, but they are all generalized by the following theorem:

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Theorem: Let $f_1,\cdots,f_k:U\to\mathbb{R}^m$, where $U\subseteq\mathbb{R}^n$, be differentiable at $a\in U$ and let $K:\left(\mathbb{R}^m\right)^k\to\mathbb{R}^s$ be multilinear then $M:U\to\mathbb{R}^s$ given by $M(x)=K(f_1(x),\cdots,f_k(x))$ is differentiable at $a$ and

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$\displaystyle D_M(a)(x)=\sum_{j=1}^{k}K\left(f_1(a),\cdots,D_{f_j}(a)(x),\cdots,f_k(a)\right)$

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Proof: Unsurprisingly we mimic the special cases we’ve already done. In particular, we define $F:\mathbb{R}^n\to\mathbb{R}^{mk}$ by

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$F(x)=(f_{1,1}(x),\cdots,f_{1,m}(x),\cdots,f_{k,1}(x),\cdots,f_{k,m}(x))$

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(where $f_{i,j}$ is the $j^{\text{th}}$ coordinate function of $f_i$). We know that $F$ is differentiable at $a$ since each of its coordinate functions is differentiable at $a$. Also, we know that $K$ is differentiable since it’s multilinear. Thus, since $M=K\circ F$ we know from the chain rule that $M$ is differentiable at $a$ and $D_M(a)=D_{K}(F(a))\circ D_F(a)$. But, as always

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$\displaystyle D_F(a)=\left(\begin{array}{ccc} & \text{Jac}_{f_{1,1}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{1,m}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{k,1}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{k,m}}(a) & \end{array}\right)$

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and we know that

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$\displaystyle D_{K}(F(a))(x)=\sum_{j=1}^{k}K\left(f_1(a),\cdots,x_j,\cdots,f_k(a)\right)$

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where $x_j\in\mathbb{R}^m$. It then follows that for any $x\in U$ one has

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

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

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As a last point in this post we generalize the ‘quotient rule’ to mappings $\mathbb{R}^n\to\mathbb{R}$. In other words:

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Theorem: Let $g:U\to\mathbb{R}$ where $U\subseteq\mathbb{R}^n$ be differentiable at $a\in U$ with $g(a)\ne 0$, then $\displaystyle \frac{1}{g}$ is differentiable at $a$ and

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$\displaystyle D_{\frac{1}{g}}(a)=\frac{-D_g(a)}{g(a)^2}$

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Proof: We begin by noting that since $g$ is continuous at $a$ we have that $g(a)\ne 0$ implies there exists an open ball $B_{\delta}(a)$ of $a$ for which $g$ is non-zero. Clearly then choosing $\|h\|<\delta$ implies that $g(a+h)\ne 0$. Consequently it makes sense to consider limits as $h\to\bold{0}$ with $\displaystyle \frac{1}{g(a+h)}$ in them. Thus, consider then that for $\|h\|<\delta$ one has that

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$\displaystyle \frac{1}{g(a+h)}-\frac{1}{g(a)}+\frac{D_g(a)(h)}{g(a)^2}=\frac{\left(g(a+h)-g(a)\right)D_g(a)(h)}{g(a)^2g(a+h)}-\frac{g(a+h)-g(a)+D_g(a)(h)}{g(a)g(a+h)}$

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so that for $\|h\|<\delta$

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\displaystyle \begin{aligned}\frac{ \displaystyle \left|\frac{1}{g(a+h)}-\frac{1}{g(a)}+\frac{D_g(a)(h)}{g(a)^2}\right|}{\|h\|} &\leqslant \frac{\left|(g(a+h)-g(a))D_g(a)(h)\right|}{\left|g(a)^2g(a+h)\right|\|h\|}+\frac{1}{\left|g(a)g(a+h)\right|}\frac{\left|g(a+h)-g(a)-D_g(a)(h)\right|}{\|h\|}\\ &\leqslant \frac{\left\|D_g(a)\right\|_\text{op}\left|g(a+h)-g(a)\right|}{\left|g(a)^2g(a+h)\right|}+\frac{1}{\left|g(a+h)g(a)\right|}\frac{\left|g(a+h)-g(a)-D_g(a)(h)\right|}{\|h\|}\end{aligned}

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(where $\|\cdot\|_\text{op}$ is the operator norm )but since both terms go to zero, the first since the numerator goes to zero while the denominator goes to something non-zero and the second because it’s a term going to a non-zero constant times a term which goes to zero by the definition of differentiability, we may conclude that $\displaystyle \frac{1}{g}$ is in fact differentiable at $a$ and has the claimed derivative.$\blacksquare$

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Using the product rule we evidently get the following corollary:

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Corollary: Let $f,g:U\to\mathbb{R}^m$ where $U\subseteq\mathbb{R}^n$ be differentiable at $a\in U$. If $g(a)\ne 0$ then $\displaystyle \frac{f}{g}$ is differentiable at $a$ and \$

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$\displaystyle D_{\frac{f}{g}}(a)=\frac{D_f(a)g(a)-D_g(a)f(a)}{g(a)^2}$

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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.