2.3Revision of the Single-Variable Derivative

A function $f: \bR \to \bR$ is said to be differentiable at $x = a$ if

exists.

Let's try to consider what it means for the graph of a function $f: \bR \to \bR$ to have a tangent line of slope $t$ at $(a, f(a))$. Let $x$ be a point close to $a$. To measure how much the value of our tangent line deviates from the value of $f$ at $x$, we can define a function

where the term in brackets is the tangent line evaluated at $x$. But graphically, this is rather crude: it's a rotated line not necessarily aligned with a nice axis on the coordinate plane. We move away from these “global coordinates” that depend on $x$ to a coordinate system centered at $(a, f(a))$.

Let's define $h := x - a$, the horizontal displacement between our chosen point and the point of tangency. We may now write $g$ in terms of $h$ as such:

This captures the exact same deviation of our tangent line from the function.

Why is this perspective useful? Because it normalizes our problem. If our candidate slope $t$ is indeed the true tangent slope of $f$ at $a$, then near $a$, the function $f$ behaves like a line of slope $t$. By subtracting the line $f(a) + th$ to create $g(h)$, we are subtracting out the entirety of this linear behavior. A curve of slope $t$ minus a line of slope $t$ leaves a remainder curve $g$ with a slope of exactly $0$.

Thus, the messy problem of checking if $f$ has a tangent of slope $t$ out at $x = a$ is entirely reduced to a much tidier problem: checking if $g$ is perfectly horizontal at the origin.

To say that $g$ is horizontal at the origin geometrically means that for any arbitrarily small slope $c > 0$, the graph of $g$ will eventually fall entirely within the wedge bounded by the lines $y = ch$ and $y = -ch$ if we zoom in close enough to the origin. Analytically, this means $|g(h)| \le c|h|$ for small enough $h$, which is exactly the definition of $g(h)$ being $o(h)$.

We can now show that $t = f'(a)$. Note that

where the numerator is exactly of the form $g(h)$, and this limit expression also agrees that $g(h)$ is $o(h)$. Thus the derivative $f'(a)$ is exactly the value of $t$ for which our tangent line is actually tangent to $f$ at $x = a$.

Let $A \subseteq \bR^n$, and let $f: A \to \bR^m$ be a mapping. $f$ is said to be differentiable at some interior point $a \in \Int(A)$ if there exists some $T_a: \bR^n \to \bR^m$ satisfying the condition that

Alternatively, $f$ is differentiable at $a$ if

where $o(h)$ is some $o(h)$ mapping.

$T_a$ is said to be the total derivative of $f$ at $a$, written $Df_a$ or $(Df)_a$. When $f$ is differentiable at $a$, the matrix $[Df_a]$ expressed in the standard basis is said to be the Jacobian of $f$ at $a$.

Recall that, for a univariate function $f: [a, b] \to \bR$, $f'$ is not well defined at $a, b$. This is because $a, b \notin \Int([a, b])$, an idea expressed in our definition of multivariate differentiability.

Let's apply this idea to some $f: A \to \bR$, where $A \subseteq \bR^2$. We say $f$ is differentiable at some interior point of $(x, y) \in \Int(A)$ if

Obviously, since $T_a$ is a linear map, we note $T_a(h, k) = ah + bk$ for some choice of values $a, b \in \bR$. Our statement of differentiability, rearranged, says that

The right-hand side is the equation of a plane in local coordinates $(h, k)$, leading us to believe that $f$ is differentiable at $(x, y)$ if, locally, $f$ behaves like a plane. This is exactly our intuition of what is a “tangent plane” for a surface that is differentiable at some point.

Let $f: A \to \bR^m$, where $A \subseteq \bR^n$, be differentiable at $a \in A$. Then the derivative of $f$ at $a$ is unique.

Suppose $f: A \to \bR^m$, where $A \subseteq \bR^m$, is differentiable at $a \in A$. Then $f$ is continuous at $a$.