DIFFERENTIATION
Section 6.1 The Derivative
6.1 THE DERIVATIVE 163
(b) The function f þg is differentiable at c, and
ð4Þ ðf þgÞ0ð Þ ¼c f0ð Þ þc g0ð Þ:c (c) (Product Rule) The function fg is differentiable at c, and ð5Þ ð Þf g 0ð Þ ¼c f0ð Þg cc ð Þ þf cð Þg0ð Þ:c
(d) (Quotient Rule)If g cð Þ 6¼0,then the function f=g is differentiable at c, and
ð6Þ f
g
0
ðcÞ ¼f0ð Þg cc ð Þ f cð Þg0ð Þc g cð Þ
ð Þ2 :
Proof. We shall prove (c) and (d), leaving (a) and (b) as exercises for the reader.
(c) Letp:¼f g; then forx2I; x6¼c, we have p xð Þ p cð Þ
xc ¼ f xð Þg xð Þ f cð Þg cð Þ xc
¼ f xð Þg xð Þ f cð Þg xð Þ þf cð Þg xð Þ f cð Þg cð Þ xc
¼ f xð Þ f cð Þ
xc g xð Þ þf cð Þ g xð Þ g cð Þ xc : Since g is continuous atc, by Theorem 6.1.2, then lim
x!cg xð Þ ¼g cð Þ. Since f andg are differentiable atc, we deduce from Theorem 4.2.4 on properties of limits that
x!clim
p xð Þ p cð Þ
xc ¼f0ð Þg cc ð Þ þf cð Þg0ð Þ:c Hencep:¼f gis differentiable atcand (5) holds.
(d) Let q:¼f=g. Since g is differentiable at c, it is continuous at that point (by Theorem 6.1.2). Therefore, sinceg cð Þ 6¼0, we know from Theorem 4.2.9 that there exists an intervalJIwithc2Jsuch thatg xð Þ 6¼0 for allx2J. Forx2J;x6¼c, we have
q xð Þ q cð Þ
xc ¼ f xð Þ=g xð Þ f cð Þ=g cð Þ
xc ¼f xð Þg cð Þ f cð Þg xð Þ g xð Þg cð ÞðxcÞ
¼ f xð Þg cð Þ f cð Þg cð Þ þf cð Þg cð Þ f cð Þg xð Þ g xð Þg cð ÞðxcÞ
¼ 1
g xð Þg cð Þ
f xð Þ f cð Þ
xc g cð Þ f cð Þ g xð Þ g cð Þ xc
: Using the continuity ofgatcand the differentiability offandgatc, we get
q0ð Þ ¼c lim
x!c
q xð Þ q cð Þ
xc ¼f0ð Þg cc ð Þ f cð Þg0ð Þc g cð Þ
ð Þ2 :
Thus,q¼f=gis differentiable atcand equation (6) holds. Q.E.D.
Mathematical Induction may be used to obtain the following extensions of the differentiation rules.
6.1.4 Corollary If f1;f2;. . .;fn are functions on an interval I toR that are differen- tiable at c2I, then:
(a) The function f1þf2þ þfn is differentiable at c and
ð7Þ ðf1þf2þ þfnÞ0ð Þ ¼c f01ð Þ þc f02ð Þ þ þc f0nð Þ:c
(b) The function f1f2 fn is differentiable at c, and
ð8Þ ðf1f2 fnÞ0ð Þ ¼c f01ð Þc f2ð Þ c fnð Þ þc f1ð Þc f02ð Þ c fnð Þc þ þf1ð Þc f2ð Þ c f0nð Þ:c
An important special case of the extended product rule (8) occurs if the functions are equal, that is,f1¼f2¼ ¼fn¼f. Then (8) becomes
ð9Þ ð Þfn 0ð Þ ¼c n f cð ð ÞÞn1f0ð Þ:c
In particular, if we take f xð Þ:¼x, then we find the derivative of g xð Þ:¼xn to be g0ð Þ ¼x nxn1; n2N. The formula is extended to include negative integers by applying the Quotient Rule 6.1.3(d).
Notation IfIRis an interval and f :I!R, we have introduced the notation f0 to denote the function whose domain is a subset ofIand whose value at a pointcis the derivative f0ð Þc offatc. There are other notations that are sometimes used forf0; for example, one sometimes writesDfforf0. Thus one can write formulas (4) and (5) in the form:
D fð þgÞ ¼DfþDg; D f gð Þ ¼ð Þ Df gþf ð Þ:Dg
Whenxis the ‘‘independent variable,’’ it is common practice in elementary courses to write df=dxforf0. Thus formula (5) is sometimes written in the form
d
dxðf xð Þg xð ÞÞ ¼ df dxð Þx
g xð Þ þf xð Þ dg dxð Þx
:
This last notation, due to Leibniz, has certain advantages. However, it also has certain disadvantages and must be used with some care.
The Chain Rule
We now turn to the theorem on the differentiation of composite functions known as the
‘‘Chain Rule.’’ It provides a formula for finding the derivative of a composite function g f in terms of the derivatives ofg andf.
We first establish the following theorem concerning the derivative of a function at a point that gives us a very nice method for proving the Chain Rule. It will also be used to derive the formula for differentiating inverse functions.
6.1.5 Caratheodory’s Theorem Let f be defined on an interval I containing the point c.
Then f is differentiable at c if and only if there exists a functionwon I that is continuous at c and satisfies
ð10Þ f xð Þ f cð Þ ¼wð Þx ðxcÞ f or x2I:
In this case, we havewð Þ ¼c f0ð Þc.
Proof. ð Þ) Iff0ð Þc exists, we can definew by
wð Þx :¼
f xð Þ f cð Þ
xc for x6¼c;x2I;
f0ð Þc for x¼c:
8<
:
The continuity ofwfollows from the fact that lim
x!cwð Þ ¼x f0ð Þc. Ifx¼c, then both sides of (10) equal 0, while ifx6¼c, then multiplication ofwð Þx byxcgives (10) for all otherx2I. 6.1 THE DERIVATIVE 165
(
ð ÞNow assume that a functionwthat is continuous atcand satisfying (10) exists. If we divide (10) byxc6¼0, then the continuity ofwimplies that
wð Þ ¼c lim
x!cwð Þ ¼x lim
x!c
f xð Þ f cð Þ xc
exists. Thereforefis differentiable atcandf0ð Þ ¼c wð Þc . Q.E.D.
To illustrate Caratheodory’s Theorem, we consider the functionfdefined byf xð Þ:¼x3 forx2R. Forc2R, we see from the factorization
x3c3¼ x2þcxþc2 xc ð Þ
thatwð Þx :¼x2þcxþc2satisfies the conditions of the theorem. Therefore, we conclude thatfis differentiable atc2Rand thatf0ð Þ ¼c wð Þ ¼c 3c2.
We will now establish the Chain Rule. Iffis differentiable atcandgis differentiable at f(c), then the Chain Rule states that the derivative of the composite functiong fatcis the productðg fÞ0ð Þ ¼c g0ðf cð ÞÞ f0ð Þc. Note that this can be written as
g f
ð Þ0¼ðg0 fÞ f0:
One approach to the Chain Rule is the observation that the difference quotient can be written, whenf xð Þ 6¼f cð Þ, as the product
g f xð ð ÞÞ g f cð ð ÞÞ
xc ¼g f xð ð ÞÞ g f cð ð ÞÞ
f xð Þ f cð Þ f xð Þ f cð Þ xc :
This suggests the correct limiting value. Unfortunately, the first factor in the product on the right is undefined if the denominatorf xð Þ f cð Þequals 0 for values ofxnearc, and this presents a problem. However, the use of Caratheodory’s Theorem neatly avoids this difficulty.
6.1.6 Chain Rule Let I, J be intervals inR, let g:I!R and f :J!R be functions such that f Jð Þ I, and let c2J. If f is differentiable at c and if g is differentiable at f(c), then the composite function g f is differentiable at c and
ð11Þ ðg fÞ0ð Þ ¼c g0ðf cð ÞÞ f0ð Þ:c
Proof. Since f0ð Þc exists, Caratheodory’s Theorem 6.1.5 implies that there exists a function w on J such that w is continuous at c and f xð Þ f cð Þ ¼wð Þx ðxcÞ for x2J, and wherewð Þ ¼c f0ð Þc. Also, sinceg0ðf cð ÞÞexists, there is a functioncdefined onIsuch thatcis continuous atd :¼f cð Þandg yð Þ g dð Þ ¼cð ÞyðydÞfory2I, where cð Þ ¼d g0ð Þd . Substitution ofy¼f xð Þandd ¼f cð Þthen produces
g f xð ð ÞÞ g f cð ð ÞÞ ¼cðf xð ÞÞðf xð Þ f cð ÞÞ ¼½ðc f xð ÞÞ wð Þx ðxcÞ for allx2J such thatf xð Þ 2I. Since the functionðc fÞ wis continuous atcand its value atcisg0ðf cð ÞÞ f0ð Þc, Caratheodory’s Theorem gives (11). Q.E.D.
Ifgis differentiable onI, iffis differentiable onJ, and iff Jð Þ I, then it follows from the Chain Rule that ðg fÞ0¼ðg0 fÞ f0, which can also be written in the form D gð fÞ ¼ðDg fÞ Df.
6.1.7 Examples (a) If f :I!R is differentiable on Iand g yð Þ:¼yn for y2R and n2N, then sinceg0ð Þ ¼y nyn1, it follows from the Chain Rule 6.1.6 that
g f
ð Þ0ð Þ ¼x g0ðf xð ÞÞ f0ð Þx for x2I:
Therefore we have ð Þfn 0ð Þ ¼x n f xð ð ÞÞn1f0ð Þx for allx2I as was seen in (9).
(b) Suppose thatf :I!Ris differentiable onIand thatf xð Þ 6¼0 andf0ð Þ 6¼x 0 forx2I. Ifh yð Þ:¼1=yfory6¼0, then it is an exercise to show thath0ð Þ ¼ y 1=y2fory2R;y6¼0.
Therefore we have 1 f
0ð Þ ¼x ðh fÞ0ð Þ ¼x h0ðf xð ÞÞf0ð Þ ¼ x f0ð Þx f xð Þ
ð Þ2 for x2I:
(c) The absolute value functiong xð Þ:¼j jx is differentiable at allx6¼0 and has derivative g0ð Þ ¼x sgnð Þx forx6¼0. (The signum function is defined in Example 4.1.10(b).) Though sgn is defined everywhere, it is not equal tog0 atx¼0 sinceg0ð Þ0 does not exist.
Now iffis a differentiable function, then the Chain Rule implies that the function g f ¼ jfj is also differentiable at all points x where f xð Þ 6¼0, and its derivative is given by
jfj0ð Þ ¼x sgnðf xð ÞÞ f0ð Þ ¼x f0ð Þx if f xð Þ>0; f0ð Þx if f xð Þ<0:
Iffis differentiable at a point cwithf cð Þ ¼0, then it is an exercise to show thatj jf is differentiable atcif and only iff0ð Þ ¼c 0. (See Exercise 7.)
For example, iff xð Þ:¼x21 forx2R, then the derivative of its absolute value f
j jð Þ ¼x x21is equal toj jf 0ð Þ ¼x sgnðx21Þ ð Þ2x forx6¼1; 1. See Figure 6.1.1 for a graph ofj jf .
(d) It will be proved later that ifS xð Þ:¼sinxandC xð Þ:¼cosxfor allx2R, then S0ð Þ ¼x cosx¼C xð Þ and C0ð Þ ¼ x sinx¼ S xð Þ
for allx2R. If we use these facts together with the definitions tanx:¼sinx
cosx; secx:¼ 1 cosx; Figure 6.1.1 The functionj jfð Þ ¼x x21.