Total Derivatives
1. Definition of Differentiability
Definition
Suppose \(U \subset {\mathbb R}^n\) is open and \(f : U \rightarrow {\mathbb R}^m\). We say that \(f\) is differentiable at \(p \in U\) when there exists a linear map \(D_p f : {\mathbb R}^n \rightarrow {\mathbb R}^m\) such that
\[ \lim_{x \rightarrow p} \frac{f(x) - f(p) - (D_p f) (x-p)}{||x-p||} = 0. \]
Theorem
If \(U \subset {\mathbb R}^n\) is open and \(f : U \rightarrow {\mathbb R}^m\) is differentiable at \(p \in U\), then the derivative is unique.
Proof
Suppose there were two linear maps \(L_1, L_2\) such that
\[ \lim_{x \rightarrow p} \frac{f(x) - f(p) - L_i (x-p)}{||x-p||} = 0. \]
for \(i = 1,2\). Taking the difference of the limits gives that
\[ \lim_{x \rightarrow p} \frac{(L_1-L_2) (x-p)}{||x-p||}\]
However, if there is any vector \(v\) such that \(L_1v \neq L_2 v\), then by letting \(x_n := p + t_n v\) for a sequence of positive \(t\)'s tending to zero, it follows that
\[ \frac{(L_1 - L_2)(x_n-p)}{||x_n-p||} = \frac{(L_1 - L_2)(v)}{||v||}\]
which does not tend to infinity as \(n \rightarrow \infty\). The sequential characterization of limits then dictates that the original limit could not possibly equal zero.
2. Differentiability, Continuity, and Partial Differentiability
Theorem
If \(U \subset {\mathbb R}^n\) is open and \(f : U \rightarrow {\mathbb R}^m\) is differentiable at \(p \in U\), then \(f\) is continuous at \(p\).
Proof
Theorem
If \(U \subset {\mathbb R}^n\) is open and \(f : U \rightarrow {\mathbb R}^m\) is differentiable at \(p \in U\), then directional derivatives of \(f\) exist in all directions and the derivative matrix of \(f\) is the Jacobian matrix
\[ \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n}
\end{bmatrix}\]
Proof
Theorem
If \(U \subset {\mathbb R}^n\) is open and \(f : U \rightarrow {\mathbb R}^m\) is partially differentiable in all coordinate directions on a neighborhood of \(p\) and the partial derivatives of \(f\) are continuous at \(p\), then \(f\) is differentiable at \(p\).
Proof
3. Algebraic Differentiation Laws
Theorem
If \(U \subset {\mathbb R}^n\) is open and \(f,g : U \rightarrow {\mathbb R}^m\) are both differentiable at \(p\), then
- \(f + g\) is differentiable at \(p\) and \(D_p (f+g) = D_p f + D_p g\)
- \(\alpha f\) is differentiable at \(p\) for any \(\alpha \in {\mathbb R}\) and \(D_p (\alpha f) = \alpha D_p f\).
- \(f \cdot g\) is differentiable at \(p\) and\[{}D_p(f \cdot g)(h){}\]\[{}= (D_p f (h)) \cdot g(p) + f(p) \cdot (D_p g(h)){}\]for all \(h \in {\mathbb R}^n\), where \(\cdot\) denotes the usual dot product on \({\mathbb R}^m\). In other words, regarding \(f\) and \(g\) as column vectors gives that\[ D_p (f \cdot g) = (g(p))^T D_p f + (f(p))^T D_p g. \]
Proof