23 Properties
Highlights of this Chapter: we prove many foundational theorems about the derivative that one sees in an early calculus course. We see how to take the derivative of scalar multiples, sums, products, quotients and compositions. We also compute - directly from the definition - the derivative of exponential functions. This leads to an important discovery: there is a unique simplest, or natural exponential, whose derivative is itself. This is the origin of \(e\) in Analysis.
From the definition, we move on to confirm the basic properties of the derivative well known and loved in introductory calculus courses. Most of these are straightforward, the only exception whose proof requires more thought than usually let on in Calculus I is the chain rule.
However before jumping in we prove one small oft-useful result often not mentioned in a calculus class, relating differentiability to continuity.
Proposition 23.1 Let \(f\) be differentiable at \(a\in\RR\). Then \(f\) is continuous at \(a\).
Proof. Since \(f\) is differentiable at \(a\), we know the limit of the difference quotient is finite \[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f^\prime(a)\] We also know that \(\lim_{x\to a}(x-a)=0\)$ So, using the limit theorems we may multiply these together and get what we want. Precisely, let \(x_n\to a\) be any sequence with \(x_n\neq a\) for all \(n\). Then we have
\[\begin{align*} 0 &= (0)(f^\prime(a))\\ &=\left(\lim x_n-a\right)\left(\lim \frac{f(x_n)-f(a)}{x_n-a}\right)\\&=\lim\left((x_n-a)\frac{f(x_n)-f(a)}{x_n-a}\right)\\ &=\lim\left(f(x_n)-f(a)\right) \end{align*}\]
Thus \(\lim (f(x_n)-a)=0\) so by the limit theorems we see \(\lim f(x_n)=a\). Since \(x_n\) was arbitrary with \(x_n\neq a\) this holds for any such sequence, we see that \(f\) is continuous at \(a\) using the sequence definition.
There is a little gap not explicitly spelled out at the end of the proof above, that we should fill in now (to assure ourselves this style of reasoning always works). We just proved that for sequences \(x_n\neq a\) the property we want holds, but continuity requires this fact for all arbitrary sequences. How do we bridge this gap? Let \(y_n\to a\) be an arbitrary sequence: then we split into the subsequences \(x_n\neq a\) and the subsequence of all terms \(=a\). If either of these is finite, we can just truncate the original sequence at a point past which all terms are of one or the other: each of these has \(\lim f(x_n)=f(a)\) so we are done. In the case that both are infinite, we just use that we have separated our sequence into a union of two subsequences, each with the same limit! Thus the overall limit exists.
23.1 Differentiation and Field Operations
Here we prove the ‘derivative laws’ of Calculus I:
Theorem 23.1 Let \(f\) be a function and \(c\in\RR\). Then if \(f\) is differentiable at a point \(a\in\RR\) so is \(cf\), and \[(cf)^\prime(a)=c\left(f^\prime(a)\right)\]
Proof. Let’s use the difference quotient with \(a+h_n\) to change things up: Let \(h_n\to 0\) be arbitrary, and we wish to compute the limit \[\lim \frac{cf(a+h_n)-cf(a)}{h_n}\] By the limit laws we can pull out the constant \(c\), and the remainder converges to \(f^\prime(a)\), as \(f\) is assumed to be differentiable at \(a\).
\[=c\lim \frac{f(a+h_n)-f(a)}{h_n}=cf^\prime(a)\]
Because this is true for all sequences \(h_n\to 0\) with \(h_n\neq 0\), the limit exists, and equals \(cf^\prime(a)\).
Exercise 23.1 Let \(f,g\) be functions which are both differentiable at a point \(a\in\RR\). Then \(f+g\) is also differentiable at \(a\), and \[(f+g)^\prime(a)=f^\prime(a)+g^\prime(a)\]
Theorem 23.2 (The Product Rule) Let \(f,g\) be functions which are both differentiable at a point \(a\in\RR\). Then \(fg\) is differentiable at \(a\) and
\[(fg)^\prime(a)=f^\prime(a)g(a)+f(a)g^\prime(a)\]
Proof. Let \(f,g\) be differentiable at \(a\in\RR\), and choose an arbitrary sequence \(a_n\to a\). Then we wish to compute
\[\lim\frac{f(a_n)g(a_n)-f(a)g(a)}{a_n-a}\]
To the numerator we add \(0=f(a_n)g(a)-f(a_n)g(a)\) and regroup with algebra:
\[=\lim \frac{f(a_n)g(a_n)-f(a_n)g(a)+f(a_n)g(a)-f(a)g(a)}{a_n-a}\] \[=\lim\frac{f(a_n)g(a_n)-f(a_n)g(a)}{a_n-a}+\frac{f(a_n)g(a)-f(a)g(a)}{a_n-a}\]
Using the limit laws, we can take each of these limits individually so long as they exist (which we will show they do). But even more, note that the first term has a common factor of \(f(a_n)\) in the numerator that can be factored out, and the second a common factor of \(g(a)\). Thus, by the limit laws, we see
\[=\left(\lim f(a_n)\right)\left(\lim\frac{g(a_n)-g(a)}{a_n-a}\right)+g(a)\left(\frac{f(a_n)-f(a)}{a_n-a}\right)\]
Because \(f\) is differentiable at \(a\), its continuous at \(a\), and so we know \(\lim f(a_n)=f(a)\). The other two limits above converge to the derivatives \(f^\prime(a)\) and \(g^\prime(a)\) respectively. Thus, alltogether we find the resulting limit to be
\[f(a)g^\prime(a)+f^\prime(a)g(a)\]
As this was the result for an arbitrary sequence \(a_n\to a\) with \(a_n\neq a\), it must be the same for all sequences, meaning the limit exists, and
\[(f\cdot g)^\prime (a)=f(a)g^\prime(a)+f^\prime(a)g(a)\]
Exercise 23.2 (The Reciprocal Rule) Let \(f\) be a function and \(a\in\RR\) be a point such that \(f(a)\neq 0\) and \(f\) is differentiable at \(a\). Then \(1/f\) is also differentiable at \(a\) and \[\left(\frac{1}{f}\right)^\prime(a)=\frac{-f^\prime(a)}{f(a)^2}\]
Theorem 23.3 (The Quotient Rule) Let \(f,g\) be a functions which are differentiable at a point \(a\in\RR\) and assume \(g(a)\neq 0\). Then the function \(f/g\) is also differentiable at \(a\) and \[\left(\frac{f}{g}\right)^\prime(a)=\frac{f^\prime(a)g(a)-f(a)g^\prime(a)}{g(a)^2}\]
Exercise 23.3 Use the Reciprocal Rule and Product Rule to prove the quotient rule.
23.2 The Chain Rule
Theorem 23.4 (The Chain Rule) If \(g(x)\) is differentiable at \(a\in\RR\) and \(f(x)\) is differentiable at \(g(a)\) then the composition \(f\circ g\) is differentiable at \(a\), with \[(f\circ g)^\prime(a)=f^\prime(g(a))g^\prime(a)\]
Proof (Wish this Worked!). We are taking the derivative at \(a\), so let \(x_n\to a\) wtih \(x_n\neq a\) be arbitrary. Then the limit defining \(\left[f(g(a))\right]^\prime\) is
\[\lim \frac{f(g(x_n))-f(g(a))}{x_n-a}\]
We multiply the numerator and denominator of this fraction by $\(g(x_n)-g(a)\) and regroup:
\[\begin{align*} \frac{f(g(x_n))-f(g(a))}{x_n-a}&= \frac{f(g(x_n))-f(g(a))}{x_n-a}\frac{g(x_n)-g(a)}{g(x_n)-g(a)}\\ &=\lim \frac{f(g(x_n))-f(g(a))}{g(x_n)-g(a)}\frac{g(x_n)-g(a)}{x_n-a} \end{align*}\]
Because \(g\) is continuous at \(a\), we know \(g(x_n)\to a\), and because \(f\) is differentiable at \(g(a)\) we recognize the first term here as the limit defining \(f^\prime\) at \(g(a)\)! Since the second term is the limit defining the derivative of \(g\), both of these exist by our assumptions, and so by the limit theorems we can compute
\[= \left(\lim \frac{f(g(x_n))-f(g(a))}{g(x_n)-g(a)}\right)\left(\lim \frac{g(x_n)-g(a)}{x_n-a}\right)\] \[ = f^\prime(g(a))g^\prime(a) \]
Unfortunately, this proof fails at one crucial step! Wile we do know that \(x_n-a\neq 0\) (in the definition of \(\lim_{x\to a}\), we only choose sequences \(x_n\to a\) with \(x_n\neq a\)) we do not know that the other denominator \(g(x_n)-g(a)\) is nonzero.
If this problem could only happen finitely many times it would be no trouble - we could just truncate the beginning of our sequence and rest assured we had not affected the value of the limit. But functions - even differentiable functions - can be pretty wild. The function \(x^2\sin(1/x)\) (from Example 22.3) ends up equaling zero infinitely often in any neighborhood of zero! So such things are a real concern.
Happily the fix - while tedious - is straightforward. It’s given below.
Exercise 23.4 We define the auxiliary function \(d(y)\) as follows:
\[d(y)=\begin{cases} \frac{f(y)-f(g(a))}{y-g(a)} & y\neq g(a)\\ f^\prime(g(a))& y=g(a) \end{cases}\]
This function equals our problematic difference quotient most of the time, but equals the quantity we want it to be when the denominator is zero.
Prove that \(d\) is continuous at \(g(c)\) and we may use \(d\) in place of the difference quotient in our computation: that for all \(x\neq a\), the following equality holds:
\[\frac{f(g(x))-f(g(a))}{x-a}=d(g(x))\frac{g(x)-g(a)}{x-a}\]
Given this, the original proof is rescued:
Proof. We are taking the derivative at \(a\), so let \(x_n\to a\) with \(x_n\neq a\) be arbitrary. Then the limit defining \(\left[f(g(a))\right]^\prime\) is (by the exercise)
\[\lim \frac{f(g(x_n))-f(g(a))}{x_n-a}=\lim d(g(x_n))\frac{g(x_n)-g(a)}{x_n-a}\]
Because \(d\) is continuous at \(g(a)\) and \(g(x_n)\to g(a)\) we know \(d(g(x_n))\to d(g(a))=f^\prime(g(a))\). And, as \(g\) is differentiable at \(a\) we know the limit of the difference quotient exists. Thus, by the limit laws we can separate them and
\[=\left(\lim d(g(x_n))\right)\left(\frac{g(x_n)-g(a)}{x_n-a}\right)=f^\prime(g(a))g^\prime(a)\]
23.3 Exponentials and Logs
In this section we look at how to find derivatives of functions which are defined not explicitly, but by functional equations. We will take the exponential as our example case; on the final project you will analyze the trigonometric functions this way.
Proposition 23.2 Let \(E(x)\) be an exponential function. Then \(E\) is differentiable on the entire real line, and \[E^\prime(x) = E^\prime(0)E(x)\]
First we show that this formula holds so long as \(E\) is actually differentiable at zero. Thus, differentiability at a single point is enough to ensure differentiability everywhere and fully determine the formula!
Proof. Let \(x\in\RR\), and \(h_n\to 0\). Then we compute \(E^\prime(x)\) by the following limit: \[E^\prime(x)=\lim \frac{E(x+h_n)-E(x)}{h_n}\]
Using the property of exponentials and the limit laws, we can factor an \(E(x)\) out of the entire numerator:
\[=\lim \frac{E(x)E(h_n)-E(x)}{h_n}=E(x)\lim \frac{E(h_n)-1}{h_n}\]
But, \(E(0)=1\) so the limit here is actually the *derivative of \(E\) at zero$!
\[E^\prime(x)=E(x)E^\prime(0)\]
Next, we tackle the slightly more subtle problem of showing that \(E\) is in fact differentiable at zero. This is tricky because all we have assumed is that \(E\) is continuous and satisfies the law of exponents: how are we going to pull differentiability out of this? The trick is two parts (1) show the right and left hand limits defining the derivative exist, and (2) show they’re equal.
Proof. STEP 1: Show that the left and right hand limits defining the derivative exist: \(E\) is convex (Exercise 15.3) so the difference quotient is monotone increasing (Proposition 13.1), and so the limit \(\lim_{x\to 0^-}\) exists (as a sup) and \(\lim_{x\to 0^+}\) exists (as an inf), Corollary 22.1.
STEP2: Now that we know each of these limits exist, let’s show they are equal using the definition:
To compute the lower limit, we can choose any sequence approaching \(0\) from below: let \(h_n\) be a positive sequence with \(h_n\to 0\), then \(-h_n\) will do:
\[\lim_{h\to 0^-}\frac{E(h)-1}{h}=\lim \frac{E(-h_n)-1}{-h_n}\] And by Exercise 15.2 we see \(E(-h_n)=1/E(h_n)\). Thus \[\begin{align*} \lim \frac{E(-h_n)-1}{-h_n}&=\lim \frac{\frac{1}{E(h_n)}-1}{-h_n}\\ &=\lim\frac{1-E(h_n)}{-h_n}\frac{1}{E{h_n}}\\ &=\lim \frac{E(h_n)-1}{h_n}\frac{1}{E(h_n)} \end{align*}\]
But, since \(E\) is continuous (by definition) and \(E(0)=1\) (Exercise 15.2) the limit theorems imply \[\lim \frac{1}{E(h_n)}=\frac{1}{\lim E(h_n)}=\frac{1}{E(\lim h_n)}=\frac{1}{E(0)}=1\] Thus, \[\begin{align*} &\lim \left(\frac{E(h_n)-1}{h_n}\frac{1}{E(h_n)}\right)\\&= \left(\lim \frac{E(h_n)-1}{h_n}\right)\left(\lim\frac{1}{E(h_n)}\right)\\ &=\lim \frac{E(h_n)-1}{h_n}\\ \end{align*}\] But this last limit evaluates exactly to the limit from above since \(h_n>0\) and \(h_n\to 0\). Stringing all of this together, we finally see \[\lim_{h\to 0^-}\frac{E(h)-1}{h}=\lim_{h\to 0^+}\frac{E(h)-1}{h}\] Thus, by Theorem 17.2 we see that since both one sided limits exist and are equal the entire limit exists: \(E\) is differentiable at \(0\).
This theorem tells us that the exponential functions have a remarkable property: they are their own derivatives, up to a constant multiple! While the functional equation alone did not provide us any means of distinguishing between different exponential functions, differentiation selects a single best, or simplest exponential out of the lot: the one where that constant multiple is just \(1\)!
Definition 23.1 We write \(\exp(x)\) for the exponential function which has \(\exp^\prime(0)=1\).
Note that by the chain rule we know such a thing exists so long as any exponential exists. If \(E(x)\) is any exponential then \(E(x/E^\prime(0))\) has derivative \(1\) at \(x=0\)!
Recalling our work with irrational exponents, we saw that if \(E\) is an exponential with \(E(1)=a\), then we may write \(E(x)=a^x\) for any \(x\in\RR\) (defined as a limit of rational exponents). So, our special exponential \(\exp\) comes with a special number as its base.
Definition 23.2 We denote by the letter \(e\) the base of the exponential \(\exp(x)\): that is, \(e=\exp(1)\), and \[\exp(x)=e^x\] Note that by definition we have \[(e^x)^\prime = e^x\]
This is the origin of the number \(e\) from the perspective of analysis! At this point in the story we do not know it’s value, but we now have a hint on how to get it: we just need to construct a means of computing the exponential function, and then plug in 1.
23.3.1 Logarithms
For every exponential, the inverse function is a logarithm (Theorem 15.2). So, \(E\) be any exponential, and \(L\) a logarithm. Then \(L(E(x))=x\), and differentiating with the chain rule yields
\[\left[L(E(x))\right]^\prime=L^\prime(E(x))E^\prime(x)=L^\prime(E(x))E(x)E^\prime(0)\]
The other side of the equality was \(x\), whose derivative is \(1\): thus
\[1=L^\prime(E(x))E(x)E^\prime(0)\] \[\implies L^\prime(E(x))=\frac{1}{E^\prime(0)E(x)}\]
Thus, \(L^\prime(-)\) is a function that takes the positive number \(E(x)\) to \(E^\prime(0)/E(x)\): it divides \(E^\prime(0)\) by its input!
Proposition 23.3 If \(L(x)\) is a logarithm function, then for some positive \(k\in\RR\) \[L^\prime(x)=\frac{1}{kx}\] (Indeed \(k=E^\prime(0)\) where \(E\) is the inverse of \(L\))
This tells us that like the exponential function, there is a natural logarithm - the one where the arbitrary constant appearing during differentiation is equal to \(1\).
Definition 23.3 The natural logarithm \(\log(x)\) is the logarithm function for which \[\log(x)^\prime = \frac{1}{x}\]
Corollary 23.1 (\(\log\) and \(\exp\) are Inverses)
Proof. Since \(\exp\) is an exponential, we know its derivative is some logarithm \(L\). But, differentiating \(L\) yields \[L^\prime(x)=\frac{1}{\exp^\prime(0)}\frac{1}{x}\]
Since \(\exp^\prime(0)=1\) by definition this says that \(L^\prime(x)=1/x\), which is the defining property of the natural logarithm. Thus \(L=\log\).
23.4 The Power Rule
Perhaps the most memorable fact from Calculus I is the power rule, that \((x^n)^\prime = nx^{n-1}\). In this short section, we prove the power level at various levels of generality, starting with natural number exponents and proceeding to arbitrary real exponents.
Proposition 23.4 (The Power Rule: Natural Number Exponents) If \(n\) is a natural number, \(x^n\) is differentiable at all real numbers and \[(x^n)^\prime = nx^{n-1}\]
Proof. This is directly proved via induction on \(n\), starting from the base case \(x^\prime =1\), which holds as if \(f(x)=x\) and \(a\in\RR\),
\[\lim_{x\to a}\frac{f(x)-a}{x-a}=\frac{x-a}{x-a}=1\]
Now, assume \((x^n)^\prime = nx^{n-1}\) and consider \(x^{n+1}\). Using the product rule, we compute the derivative of \(x^{n+1}=xx^{n}\)
\[\begin{align*} (xx^n)^\prime &= (x)^\prime x^n+x (x^n)^\prime\\ &= 1 x^n + x (nx^{n-1})\\ &= x^n+n x^n\\ &=(n+1)x^{n+1} \end{align*}\]
Exercise 23.5 (The Power Rule: Integer Exponents) Let \(n\in\ZZ\) and consider the function \(x^n\) (which is defined as \(1/x^{|n|}\) when \(n<0\)). Then \(x^n\) is differentiable at all \(x\neq 0\) and \[(x^n)^\prime = nx^{n-1}\]
Using this, we can extend what we know to rational exponents:
Proposition 23.5 (The Power Rule: Rational Exponents) Let \(r=p/q\) be any rational number and \(f(x)=x^r\). Then \(f\) is differentiable for all \(x>0\) and \[f^\prime(x)=rx^{r-1}\]
Proof. Let \(r=p/q\) where without loss of generality \(p,q\neq 0\) and \(q>1\) (as if \(q=1\) we are in the integer exponent case). Then let \(f(x)=x^{p/q}\), and note that \(f(x)^q = x^p\). Then we can differentiate both sides of this inequality:
\[\left [f(x)^q\right]^\prime=qf(x)^{q-1}f^\prime(x)\] \[\left[x^p\right]^\prime = px^{p-1}\]
Equating these gives \(qf(x)^{q-1}f^\prime(x)=px^{p-1}\), and solving for \(f^\prime\):
\[f^\prime(x)=\frac{px^{p-1}}{qf(x)^{q-1}}\]
Using that \(f(x)=x^{p/q}\) we can simplify the right hand side further:
\[f^\prime(x)=\frac{px^{p-1}}{q (x^{p/q})^{q-1}}=\frac{px^{p-1}}{q x^{p\frac{q-1}{a}}}=\frac{p}{q} x^{(p-1)-p\frac{q-1}{q}}\]
This exponent simplifies as expected, yielding \[f^\prime(x)=\tfrac{p}{q} x^{\frac{p}{q}-1}\]
Now that we know the power rule for all rational exponents, it is time to consider arbitrary real exponents, recalling that we define \(x^a\) as a limit of rational exponents.
Theorem 23.5 (\(\bigstar\) The General Power Rule) If \(a\in\RR\) and \(f(x)=x^a\). Then \(f\) is differentiable for all \(x>0\), and \[(x^a)^\prime = ax^{a-1}\]
Exercise 23.6 Recall the definition of \(x^a\) for irrational \(a\) is \(\lim x^{a_n}\) for \(a_n\) a sequence of rational numbers converging to \(a\). Use this definition to attempt a proof of the general power rule, by computing
\[(x^a)^\prime = \left(\lim x^{a_n}\right)^\prime\]
In your proof, you will end up getting stuck at a point where you need to interchange two limits: point out where this happens, and then show that if you are justified in interchanging the limits, that the generalized power rule holds.
Alternatively, we can give a complete proof of the power rule using exponentials and logarithms.
Proof. Let \(\exp\) be the natural exponential, and \(\log\) be the natural log. Then \(\exp(\log(x))=x\), and so \(\exp(\log(x^a))=x^a\). Using the property of logarithms and powers (Corollary 15.2) this simplifies
\[x^n=\exp(\log(x^a))=\exp(a\log(x))\]
By the chain rule,
\[\begin{align*} \left[\exp(a\log(x))\right]^\prime &= \exp(a\log(x))\left[a\log(x)\right]^\prime\\ &= \exp(a\log(x)) a \log^\prime(x)\\ &= \exp(a\log(x))a\frac{1}{x} \end{align*}\]
But, recalling that \(\exp(a\log(x))=\exp(\log(x^a))=x^a\) this simplifies to
\[=x^a a\frac{1}{x}=ax^{a-1}\]