The General Differentiation Theorems, via Limits

Section 14.2 The General Differentiation Theorems, via Limits

. . . one way in math to take care of destabilizing problems is to legislate them out of existence . . . by loading theorems with stipulations and exclusions designed to head off crazy results.
―David Foster Wallace
⁴
https://en.wikipedia.org/wiki/David_Foster_Wallace
(1962—2008)

Since we will now be proving the the differentiation rules rigorously we will call them what they really are: Theorems. Because limits are much less intuitive than differentials we’ll want to be as efficient as possible when using them. The sooner we can build up some tools to make things easier, the better.

Also, in this section we will add a new differentiation rule (theorem): The Chain Rule. Or rather, we will give a name to an already familiar technique and elevate it’s status by providing a formal proof. Proving the Chain, Product, and Quotient Differentiation Rules using limits will require a good deal of cleverness. These proofs will also uncover some unexpected subtleties along the way.

Before we begin there is one more point that needs to be clear. Because differentiation is now defined via a limit and limits are defined at a point we can only differentiate a function at a point. We usually say that limit evaluation and differentiability are local properties. If we don’t specify the “at \(x\)” the convention is that the function is differentiable at every point in its domain.

The proof of the Constant Rule is possibly the simplest proof involving limits in existence. We provide the proof below in order To display the formalism of a proof using limits. You should use it as a guide for the exercises that follow it.

Theorem 14.2.1. The Constant Rule for Differentiation.

If \(L\) is some number and \(f(x)=L\) for all real values of \(x\) near (on an open interval around) \(L\text{,}\) then \(f^\prime(x)=0\) at every real number \(x\) near (on the same open interval) \(L\text{.}\)

Proof.

By equation (13.3)

\begin{equation*} f^\prime(x) =\limit{h}{0}{\frac{f(x+h)-f(x)}{h}}. \end{equation*}

But \(f(x+h)=L=f(x)\) so

\begin{equation*} f^\prime(x) =\limit{h}{0}{\frac{L-L}{h}}=\limit{h}{0}{\frac{0}{h}}=0. \end{equation*}

The proofs of the Sum, and Constant Multiple Differentiation Rules are all completely straightforward so we will leave them as exercises for you.

Theorem 14.2.2. The Sum Rule for Differentiation.

If \(\alpha(x)\) and \(\beta(x)\) are differentiable at \(x\) and \(f(x)=\alpha(x)+\beta(x)\text{,}\) then \(f(x)\) is also differentiable at \(x\) and

\begin{equation*} f^\prime(x)=\alpha^\prime(x)+ \beta^\prime(x). \end{equation*}

Problem 14.2.3.

Use Definition 13.2.3 to prove the Sum Rule for Differentiation.

Recall that when we first established the General Differentiation Rules using differentials in Chapter 4 we said that the Constant Multiple, Power and Quotient Rules for differentiation were just conveniences because they depend on the other rules. This is still true of course, which means that we don’t have to prove any of them using limits. However is is straightforward to prove the Constant Multiple Rule directly from the limit definition of a derivative.

Theorem 14.2.4. The Constant Multiple Rule for Differentiation.

If \(f(x)\) is differentiable at \(x\) and \(K\) is a constant then \(\alpha(x)=Kf(x) \) is also differentiable and

\begin{equation*} \alpha^\prime(x)=Kf^\prime(x). \end{equation*}

Problem 14.2.5.

Use the Definition 13.2.3 to prove the Constant Multiple Rule.

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