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The Derivative of a Function

Module by: Lawrence Baggett. E-mail the author

Summary: Now begins what is ordinarily thought of as the first main subject of calculus, the derivative. A definition of differentiable and some important theorems concerning derivatives, such as the chain rule, are included.

Now begins what is ordinarily thought of as the first main subject of calculus, the derivative.

Definition 1:

Let SS be a subset of R,R, let f:S→Cf:S→C be a complex-valued function (of a real variable), and let cc be an element of the interior of S.S. We say that ff is differentiable at c if

lim h → 0 f ( c + h ) - f ( c ) h lim h → 0 f ( c + h ) - f ( c ) h
(1)

exists. (Here, the number hh is a real number.)

Analogously, let SS be a subset of C,C, let f:S→Cf:S→C be a complex-valued function (of a complex variable), and let cc be an element of the interior of S.S. We say that ff is differentiable at c if

lim h → 0 f ( c + h ) - f ( c ) h lim h → 0 f ( c + h ) - f ( c ) h
(2)

exists. (Here, the number hh is a complex number.)

If f:S→Cf:S→C is a function either of a real variable or a complex variable, and if S'S' denotes the subset of SS consisting of the points cc where ff is differentiable, we define a function f':S'→Cf':S'→C by

f ' ( x ) = lim h → 0 f ( x + h ) - f ( x ) h . f ' ( x ) = lim h → 0 f ( x + h ) - f ( x ) h .
(3)

The function f'f' is called the derivative of f.f.

A continuous function f:[a,b]→Cf:[a,b]→C that is differentiable at each point x∈(a,b),x∈(a,b), and whose the derivative f'f' is continuous on (a,b),(a,b), is called a smooth function on [a,b].[a,b]. If there exists a partition {a=x0<x1<...<xn=b}{a=x0<x1<...<xn=b} of [a,b][a,b] such that ff is smooth on each subinterval [xi-1,xi],[xi-1,xi], then ff is called piecewise smooth on [a,b].[a,b].

Higher order derivatives are defined inductively. That is, f''f'' is the derivative of f',f', and so on. We use the symbol f(n)f(n) for the nnth derivative of f.f.

REMARK In the definition of the derivative of a function f,f, we are interested in the limit, as hh approaches 0, not of ff but of the quotient q(h)=f(c+h)-f(c)h.q(h)=f(c+h)-f(c)h. Notice that 0 is not in the domain of the function q,q, but 0 is a limit point of that domain. This is the reason why we had to make such a big deal above out of the limit of a function. The function qq is often called the differential quotient.

REMARK As mentioned in (Reference), we are often interested in solving for unknowns that are functions. The most common such problem is to solve a differential equation. In such a problem, there is an unknown function for which there is some kind of relationship between it and its derivatives. Differential equations can be extremely complicated, and many are unsolvable. However, we will have to consider certain relatively simple ones in this chapter, e.g., f'=f,f'=f,f'=-f,f'=-f, and f''=±f.f''=±f.

There are various equivalent ways to formulate the definition of differentiable, and each of these ways has its advantages. The next theorem presents one of those alternative ways.

Theorem 1

Let cc belong to the interior of a set SS (either in RR or in CC), and let f:S→Cf:S→C be a function. Then the following are equivalent.

  1.  ff is differentiable at c.c. That is,
    limh→0f(c+h)-f(c)hexists.limh→0f(c+h)-f(c)hexists.
    (4)
  2. lim x → c f ( x ) - f ( c ) x - c exists. lim x → c f ( x ) - f ( c ) x - c exists.
    (5)
  3. There exists a number LL and a function θθ such that the following two conditions hold:
    f(c+h)-f(c)=Lh+θ(h)f(c+h)-f(c)=Lh+θ(h)
    (6)
    and
    limh→0θ(h)h=0.limh→0θ(h)h=0.
    (7)
    In this case, LL is unique and equals f'(c),f'(c), and the function θθ is unique and equals f(c+h)-f(c)-f'(c)h.f(c+h)-f(c)-f'(c)h.

Proof

That (1) and (2) are equivalent follows from (Reference) by writing xx as c+h.c+h.

Suppose next that ff is differentiable at c,c, and define

L = f ' ( c ) = lim h → 0 f ( c + h ) - f ( c ) h . L = f ' ( c ) = lim h → 0 f ( c + h ) - f ( c ) h .
(8)

Set

θ ( h ) = f ( c + h ) - f ( c ) - f ' ( c ) h . θ ( h ) = f ( c + h ) - f ( c ) - f ' ( c ) h .
(9)

Then clearly

f ( c + h ) - f ( c ) = L h + θ ( h ) , f ( c + h ) - f ( c ) = L h + θ ( h ) ,
(10)

which is Equation 6. Also

| θ ( h ) h | = | f ( c + h ) - f ( c ) - f ' ( c ) h h | = | f ( c + h ) - f ( c ) h - f ' ( c ) | , | θ ( h ) h | = | f ( c + h ) - f ( c ) - f ' ( c ) h h | = | f ( c + h ) - f ( c ) h - f ' ( c ) | ,
(11)

which tends to 0 as hh approaches 0 because ff is differentiable at c.c. Hence, we have established Equation 6 and Equation 7, showing that (1) implies (3).

Finally, suppose there is a number LL and a function θθ satisfying Equation 6 and Equation 7. Then

f ( c + h ) - f ( c ) h = L + θ ( h ) h , f ( c + h ) - f ( c ) h = L + θ ( h ) h ,
(12)

which converges to LL as hh approaches 0 by Equation 7 and part (2) of (Reference). Hence, L=f'(c),L=f'(c), and so θ(h)=f(c+h)-f(c)-f'(c)h.θ(h)=f(c+h)-f(c)-f'(c)h. Therefore, (3) implies (1), and the theorem is proved.

REMARK Though it seems artificial and awkward, Condition (3) of this theorem is very convenient for many proofs. One should remember it.

Exercise 1

  1. What is the domain of the function θθ of condition (3) in the preceding theorem? Is 0 in this domain? Are there any points in the interior of this domain?
  2. Let LL and θθ be as in part (3) of the preceding theorem. Prove that, given an ϵ>0ϵ>0 there exists a δ>0δ>0 such that if |h|<δ|h|<δ then |θ(h)|<ϵ|h|.|θ(h)|<ϵ|h|.

Theorem 2

If f:S→Cf:S→C is a function, either of a real variable or a complex variable, and if ff is differentiable at a point cc of S,S, then ff is continuous at c.c. That is, differentiability implies continuity.

Proof

We are assuming that limh→0(f(c+h)-f(c))/h=L.limh→0(f(c+h)-f(c))/h=L. Hence, there exists a positive number δ0δ0 such that |f(c+h)-f(c)h-L|<1|f(c+h)-f(c)h-L|<1 if |h|<δ0,|h|<δ0, implying that |f(c+h)-f(c)|<|h|(|L|+1)|f(c+h)-f(c)|<|h|(|L|+1)whenever |h|<δ0.|h|<δ0. So, if ϵ>0ϵ>0 is given, let δδ be the minimum of δ0δ0 and ϵ/(|L|+1).ϵ/(|L|+1). If y∈Sy∈S and |y-c|<δ,|y-c|<δ, then, thinking of yy as being c+h,c+h,

| f ( y ) - f ( c ) | = | f ( c + h ) - f ( c ) | < | h | ( | L | + 1 ) = | y - c | ( | L | + 1 ) < ϵ . | f ( y ) - f ( c ) | = | f ( c + h ) - f ( c ) | < | h | ( | L | + 1 ) = | y - c | ( | L | + 1 ) < ϵ .
(13)

(Every yy can be written as c+hc+h for some h,h, and |y-c|=|h|.|y-c|=|h|.)

Exercise 2

Define f(z)=|z|f(z)=|z| for z∈C.z∈C.

  1. Prove that ff is continuous at every point of C.C.
  2. Show that, if ff is differentiable at a point c,c, then f'(c)=0.f'(c)=0. HINT: Using part (b) of (Reference), evaluate f'(c)f'(c) in the following two ways.
    f'(c)=limn→∞|c+1n|-|c|1nf'(c)=limn→∞|c+1n|-|c|1n
    (14)
    and
    f'(c)=limn→∞|c+in|-|c|in.f'(c)=limn→∞|c+in|-|c|in.
    (15)
    Show that the only way these two limits can be equal is for them to be 0.
  3. Conclude that ff is not differentiable anywhere. Indeed, if it were, what would the function θθ have to be, and why wouldn't it satisfy Equation 7?
  4. Suppose f:R→Rf:R→R is the function of a real variable that is defined by f(x)=|x|.f(x)=|x|. Show that ff is differentiable at every point x≠0.x≠0. How does this result not contradict part (c)?

The following theorem generalizes the preceding exercise.

Theorem 3

Suppose f:S→Rf:S→R is a real-valued function of a complex variable, and assume that ff is differentiable at a point c∈S.c∈S. Then f'(c)=0.f'(c)=0. That is, every real-valued, differentiable function ff of a complex variable satisfies f'(c)=0f'(c)=0 for all cc in the domain of f'.f'.

Proof

We compute f'(c)f'(c) in two ways.

f ' ( c ) = lim n f ( c + 1 n ) - f ( c ) 1 n is a real number. . f ' ( c ) = lim n f ( c + 1 n ) - f ( c ) 1 n is a real number. .
(16)
f ' ( c ) = lim n f ( c + i n ) - f ( c ) i n is a purely imaginary number. f ' ( c ) = lim n f ( c + i n ) - f ( c ) i n is a purely imaginary number.
(17)

Hence, f'(c)f'(c) must be 0, as claimed.

REMARK This theorem may come as a surprise, for it shows that there are very few real-valued differentiable functions of a complex variable. For this reason, whenever f:S→Rf:S→R is a real-valued, differentiable function, we will presume that ff is a function of a real variable; i.e., that the domain S⊆R.S⊆R.

Evaluating limh→0q(h)limh→0q(h) in the two different ways, hh real, and hh pure imaginary, led to the proof of the last theorem. It also leads us to make definitions of what are called “partial derivatives” of real-valued functions whose domains are subsets of C≡R2.C≡R2. As the next exercise will show, the theory of partial derivatives of real-valued functions is a much richer theory than that of standard derivatives of real-valued functions of a single complex variable.

Definition 2:

Let f:S→Rf:S→R be defined on a set S⊆C≡R2,S⊆C≡R2, and let c=(a,b)=++bic=(a,b)=++bi be a point in the interior of S.S. We define the partial derivative of f with respect to x at the point c=(a,b)c=(a,b) by the formula

t i a l f t i a l x ( a , b ) = lim h → 0 f ( a + h , b ) - f ( a , b ) h , t i a l f t i a l x ( a , b ) = lim h → 0 f ( a + h , b ) - f ( a , b ) h ,
(18)

and the partial derivative of f with respect to y at c=(a,b)c=(a,b) by the formula

t i a l f t i a l y ( a , b ) = lim h → 0 f ( a , b + h ) - f ( a , b ) h , t i a l f t i a l y ( a , b ) = lim h → 0 f ( a , b + h ) - f ( a , b ) h ,
(19)

whenever these limits exist. (In both these limits, the variable hh is a real variable.)(

It is clear that the partial derivatives of a function arise when we fix either the real part of the variable or the imaginary part of the variable to be a constant, and then consider the resulting function of the other (real) variable. We will see in Exercise 3 that there is a definite difference between a function's being differentiable at a point c=(a+bi)c=(a+bi) in the complex plane CC versus its having partial derivatives at the point (a,b)(a,b) in R2.R2.

Exercise 3

  1. Suppose ff is a complex-valued function of a complex variable, and assume that both the real and imaginary parts of ff are differentiable at a point c.c. Show that ff is differentiable at cc and that f'(c)=0.f'(c)=0.
  2. Let f=u+ivf=u+iv be a complex-valued function of a complex variable that is differentiable at a point c.c. Prove that both partial derivatives of uu and vv exist at c=(a,b),c=(a,b), and in fact that
    tialutialx(c)+itialvtialx(c)=f'(c)tialutialx(c)+itialvtialx(c)=f'(c)
    (20)
    and
    tialutialy(c)+itialvtialy(c)=if'(c).tialutialy(c)+itialvtialy(c)=if'(c).
    (21)
  3. Define a complex-valued function ff on C≡R2C≡R2 by f(z)=f(x+iy)=x-iy.f(z)=f(x+iy)=x-iy. Write f=u+iv,f=u+iv, and show that both partial derivatives of uu and vv exist at every point, but that ff is not a differentiable function of the complex variable z.z.

The next theorem is, in part, what we call in calculus the “differentiation formulas.”

Theorem 4

Let ff and gg be functions (either of a real variable or a complex variable), which are both differentiable at a point c.c. Let aa and bb be complex numbers. Then:

  1.  af+bgaf+bg is differentiable at c,c, and (af+bg)'(c)=af'(c)+bg'(c).(af+bg)'(c)=af'(c)+bg'(c).
  2.  (Product Formula) fgfg is differentiable at c,c, and (fg)'(c)=f'(c)g(c)+f(c)g'(c).(fg)'(c)=f'(c)g(c)+f(c)g'(c).
  3.  (Quotient Formula) f/gf/g is differentiable at cc (providing that g(c)≠0g(c)≠0), and
    (fg)'(c)=g(c)f'(c)-f(c)g'(c)(g(c))2.(fg)'(c)=g(c)f'(c)-f(c)g'(c)(g(c))2.
    (22)
  4. If f=u+ivf=u+iv is a complex-valued function, then ff is differentiable at a point cc if and only if uu and vv are differentiable at c,c, and f'(c)=u'(c)+iv'(c).f'(c)=u'(c)+iv'(c).

Proof

We prove part (2) and leave parts (1), (3), and (4) for the exercises. We have

lim h → 0 ( f g ) ( c + h ) - ( f g ) ( c ) h = lim h → 0 f ( c + h ) g ( c + h ) - f ( c ) g ( c ) h = lim h → 0 f ( c + h ) g ( c + h ) - f ( c ) g ( c + h ) h + lim h → 0 f ( c ) g ( c + h ) - f ( c ) g ( c ) h = lim h → 0 f ( c + h ) - f ( c ) h lim h → 0 g ( c + h ) + lim h → 0 f ( c ) lim h → 0 g ( c + h ) - g ( c ) h = f ' ( c ) g ( c ) + f ( c ) g ' ( c ) , lim h → 0 ( f g ) ( c + h ) - ( f g ) ( c ) h = lim h → 0 f ( c + h ) g ( c + h ) - f ( c ) g ( c ) h = lim h → 0 f ( c + h ) g ( c + h ) - f ( c ) g ( c + h ) h + lim h → 0 f ( c ) g ( c + h ) - f ( c ) g ( c ) h = lim h → 0 f ( c + h ) - f ( c ) h lim h → 0 g ( c + h ) + lim h → 0 f ( c ) lim h → 0 g ( c + h ) - g ( c ) h = f ' ( c ) g ( c ) + f ( c ) g ' ( c ) ,
(23)

where we have used (Reference), Theorem 1, and Theorem 2.

Exercise 4

  1. Prove parts (1), (3), and (4) of Theorem 4.
  2. If ff and gg are real-valued functions that are differentiable at a point c,c, what can be said about the differentiability of max(f,g)?max(f,g)?
  3. Let ff be a constant function f(z)≡k.f(z)≡k. Prove that ff is differentiable everywhere and that f'(z)=0f'(z)=0 for all z.z.
  4. Define a function ff by f(z)=z.f(z)=z. Prove that ff is differentiable everywhere and that f'(z)=1f'(z)=1 for all z.z.
  5. Verify the usual derivative formulas for polynomial functions: If p(z)=∑k=0nakzk,p(z)=∑k=0nakzk, then p'(z)=∑k=1nkakzk-1.p'(z)=∑k=1nkakzk-1.

What about power series functions? Are they differentiable functions? If so, are their derivatives again power series functions? In fact, everything works as expected.

Theorem 5

Let ff be a power series function f(z)=∑n=0∞anznf(z)=∑n=0∞anzn having radius of convergence r>0.r>0. Then ff is differentiable at each point zz in its open disk Br(0)Br(0) of convergence, and

f ' ( z ) = ∑ n = 0 ∞ n a n z n - 1 = ∑ n = 1 ∞ n a n z n - 1 . f ' ( z ) = ∑ n = 0 ∞ n a n z n - 1 = ∑ n = 1 ∞ n a n z n - 1 .
(24)

Proof

The proof will use part (3) of Theorem 1. Fix an zz with |z|<r.|z|<r. Choose r'r' so that |z|<r'<r,|z|<r'<r, and write αα for r'-|z|,r'-|z|, i.e., |z|+α=r'.|z|+α=r'. Note first that the infinite series ∑n=0∞|an|r'n∑n=0∞|an|r'n converges to a positive number we will call M.M. Also, from the Cauchy-Hadamard Formula, we know that the power series function ∑nanwn∑nanwn has the same radius of convergence as does f,f, and hence the infinite series ∑nanzn-1∑nanzn-1 converges to a number we will denote by L.L. We define a function θθ by θ(h)=f(z+h)-f(z)-Lhθ(h)=f(z+h)-f(z)-Lh from which it follows immediately that

f ( z + h ) - f ( z ) = L h + θ ( h ) , f ( z + h ) - f ( z ) = L h + θ ( h ) ,
(25)

which establishes Equation 6. To complete the proof that ff is differentiable at z,z, it will suffice to establish Equation 7, i.e., to show that

lim h → 0 θ ( h ) h = 0 . lim h → 0 θ ( h ) h = 0 .
(26)

That is, given ϵ>0ϵ>0 we must show that there exists a δ>0δ>0 such that if 0<|h|<δ0<|h|<δ then

| θ ( h ) / h | = | f ( z + h ) - f ( z ) h - L | < ϵ . | θ ( h ) / h | = | f ( z + h ) - f ( z ) h - L | < ϵ .
(27)

Assuming, without loss of generality, that |h|<α,|h|<α, we have that

| f ( z + h ) - f ( z ) h - L | = | ∑ n = 0 ∞ a n ( z + h ) n - ∑ n = 0 ∞ a n z n h - L | = | ∑ n = 0 ∞ a n ( ∑ k = 0 n n k z n - k h k ) - ∑ n = 0 ∞ a n z n h - L | = | ∑ n = 0 ∞ a n ( ( ∑ k = 0 n n k z n - k h k ) - z n ) h - L | = | ∑ n = 1 ∞ a n ( ∑ k = 1 n n k z n - k h k ) h - L | = | ∑ n = 1 ∞ a n ( ∑ k = 1 n n k z n - k h k - 1 ) - ∑ n = 1 ∞ n a n z n - 1 | = | ∑ n = 1 ∞ a n ( ∑ k = 1 n n k z n - k h k - 1 ) - ∑ n = 1 ∞ n 1 a n z n - 1 | = | ∑ n = 2 ∞ a n ( ∑ k = 2 n n k z n - k h k - 1 ) | ≤ ∑ n = 2 ∞ ∑ k = 2 n | a n | n k | z | n - k | h | k - 1 ≤ | h | ∑ n = 2 ∞ | a n | ∑ k = 2 n n k | z | n - k | h | k - 2 ≤ | h | ∑ n = 2 ∞ | a n | ∑ k = 2 n n k | z | n - k | α | k - 2 ≤ | h | 1 α 2 ∑ n = 0 ∞ | a n | ∑ k = 0 n n k | z | n - k α k = | h | 1 α 2 ∑ n = 0 ∞ | a n | ( | z | + α ) n = | h | 1 α 2 ∑ n = 0 ∞ | a n | r ' n = | h | M α 2 , | f ( z + h ) - f ( z ) h - L | = | ∑ n = 0 ∞ a n ( z + h ) n - ∑ n = 0 ∞ a n z n h - L | = | ∑ n = 0 ∞ a n ( ∑ k = 0 n n k z n - k h k ) - ∑ n = 0 ∞ a n z n h - L | = | ∑ n = 0 ∞ a n ( ( ∑ k = 0 n n k z n - k h k ) - z n ) h - L | = | ∑ n = 1 ∞ a n ( ∑ k = 1 n n k z n - k h k ) h - L | = | ∑ n = 1 ∞ a n ( ∑ k = 1 n n k z n - k h k - 1 ) - ∑ n = 1 ∞ n a n z n - 1 | = | ∑ n = 1 ∞ a n ( ∑ k = 1 n n k z n - k h k - 1 ) - ∑ n = 1 ∞ n 1 a n z n - 1 | = | ∑ n = 2 ∞ a n ( ∑ k = 2 n n k z n - k h k - 1 ) | ≤ ∑ n = 2 ∞ ∑ k = 2 n | a n | n k | z | n - k | h | k - 1 ≤ | h | ∑ n = 2 ∞ | a n | ∑ k = 2 n n k | z | n - k | h | k - 2 ≤ | h | ∑ n = 2 ∞ | a n | ∑ k = 2 n n k | z | n - k | α | k - 2 ≤ | h | 1 α 2 ∑ n = 0 ∞ | a n | ∑ k = 0 n n k | z | n - k α k = | h | 1 α 2 ∑ n = 0 ∞ | a n | ( | z | + α ) n = | h | 1 α 2 ∑ n = 0 ∞ | a n | r ' n = | h | M α 2 ,
(28)

so that if δ=ϵ/Mα2,δ=ϵ/Mα2, then |θ(h)/h|<ϵ,|θ(h)/h|<ϵ, whenever |h|<δ,|h|<δ, as desired.

REMARK Theorem 5 shows that indeed power series functions are differentiable, and in fact their derivatives can be computed, just like polynomials, by differentiating term by term. This is certainly a result we would have hoped was true, but the proof is not trivial.

The next theorem, the Chain Rule, is another nontrivial one. It deals with the differentiability of the composition of two differentiable functions. Again, the result is what we would have wanted, the composition of two differentiable functions is itself differentiable, but the argument required to prove it is tricky.

Theorem 6: Chain Rule

Let f:S→Cf:S→C be a function, and assume that ff is differentiable at a point c.c. Suppose g:T→Cg:T→C is a function, that T⊆C,T⊆C, that the number f(c)∈T,f(c)∈T, and that gg is differentiable at f(c).f(c). Then the composition g∘fg∘f is differentiable at cc and

( g ∘ f ) ' ( c ) = g ' ( f ( c ) ) f ' ( c ) . ( g ∘ f ) ' ( c ) = g ' ( f ( c ) ) f ' ( c ) .
(29)

Proof

Using part (3) of Theorem 1, write

g ( f ( c ) + k ) - g ( f ( c ) ) = L g k + θ g ( k ) g ( f ( c ) + k ) - g ( f ( c ) ) = L g k + θ g ( k )
(30)

and

f ( c + h ) - f ( c ) = L f h + θ f ( h ) . f ( c + h ) - f ( c ) = L f h + θ f ( h ) .
(31)

We know from that theorem that Lg=g'(f(c))Lg=g'(f(c)) and Lf=f'(c).Lf=f'(c). And, we also know that

lim k → 0 θ g ( k ) k = 0 and lim h → 0 θ f ( h ) h = 0 . lim k → 0 θ g ( k ) k = 0 and lim h → 0 θ f ( h ) h = 0 .
(32)

Define a function k(h)=f(c+h)-f(c).k(h)=f(c+h)-f(c). Then, by Theorem 2, we have that limh→0k(h)=0.limh→0k(h)=0. We will show that g∘fg∘f is differentiable at cc by showing that there exists a number LL and a function θθ satisfying the two conditions of part (3) of Theorem 1. Thus, we have that

g ∘ f ( c + h ) - g ∘ f ( c ) = g ( f ( c + h ) ) - g ( f ( c ) ) = g ( f ( c ) + k ( h ) ) - g ( f ( c ) ) = L g k ( h ) + θ g ( k ( h ) ) = L g ( f ( c + h ) - f ( c ) ) + θ g ( k ( h ) ) = L g ( L f h + θ f ( h ) ) + θ g ( k ( h ) ) = L g L f h + L g θ f ( h ) + θ g ( k ( h ) ) . g ∘ f ( c + h ) - g ∘ f ( c ) = g ( f ( c + h ) ) - g ( f ( c ) ) = g ( f ( c ) + k ( h ) ) - g ( f ( c ) ) = L g k ( h ) + θ g ( k ( h ) ) = L g ( f ( c + h ) - f ( c ) ) + θ g ( k ( h ) ) = L g ( L f h + θ f ( h ) ) + θ g ( k ( h ) ) = L g L f h + L g θ f ( h ) + θ g ( k ( h ) ) .
(33)

We define L=Lglf=g'(f(c))f'(c),L=Lglf=g'(f(c))f'(c), and we define the function θθ by

θ ( h ) = L g θ f ( h ) + θ g ( k ( h ) ) . θ ( h ) = L g θ f ( h ) + θ g ( k ( h ) ) .
(34)

By our definitions, we have established Equation 6

g ∘ f ( c + h ) - g ∘ f ( c ) = L h + θ ( h ) , g ∘ f ( c + h ) - g ∘ f ( c ) = L h + θ ( h ) ,
(35)

so that it remains to verify Equation 7.

We must show that, given ϵ>0,ϵ>0, there exists a δ>0δ>0 such that if 0<|h|<δ0<|h|<δ then |θ(h)/h|<ϵ.|θ(h)/h|<ϵ. First, choose an ϵ'>0ϵ'>0 so that

| L g | ϵ ' + | L f | ϵ ' + ϵ ' 2 < ϵ ( 4 . 3 ) . | L g | ϵ ' + | L f | ϵ ' + ϵ ' 2 < ϵ ( 4 . 3 ) .
(36)

Next, using part (b) of Exercise 1, choose a δ'>0δ'>0 such that if |k|<δ'|k|<δ' then |θg(k)|<ϵ'|k|.|θg(k)|<ϵ'|k|. Finally, choose δ>0δ>0 so that if 0<|h|<δ,0<|h|<δ, then the following two inequalities hold. |k(h)|<δ'|k(h)|<δ' and |θf(h)|<ϵ'|h|.|θf(h)|<ϵ'|h|. The first can be satisfied because ff is continuous at c,c, and the second is a consequence of part (b) of Exercise 1. Then: if 0<|h|<δ,0<|h|<δ,

| θ ( h ) | = | L g θ f ( h ) + θ g ( k ( h ) ) | ≤ | L g | | θ f ( h ) | + | θ g ( k ( h ) ) | < | L g | ϵ ' | h | + ϵ ' | k ( h ) | = | L g | ϵ ' | h | + ϵ ' | f ( c + h ) - f ( c ) | = | L g | ϵ ' | h | + ϵ ' | L f h + θ f ( h ) | ≤ | L g | ϵ ' | h | + ϵ ' | L f | | h | + ϵ ' | θ f ( h ) | < | L g | ϵ ' | h | + ϵ ' | L f | | h | + ϵ ' ϵ ' | h | = ( | L g | ϵ ' + | L f | ϵ ' + ϵ ' 2 ) | h | , | θ ( h ) | = | L g θ f ( h ) + θ g ( k ( h ) ) | ≤ | L g | | θ f ( h ) | + | θ g ( k ( h ) ) | < | L g | ϵ ' | h | + ϵ ' | k ( h ) | = | L g | ϵ ' | h | + ϵ ' | f ( c + h ) - f ( c ) | = | L g | ϵ ' | h | + ϵ ' | L f h + θ f ( h ) | ≤ | L g | ϵ ' | h | + ϵ ' | L f | | h | + ϵ ' | θ f ( h ) | < | L g | ϵ ' | h | + ϵ ' | L f | | h | + ϵ ' ϵ ' | h | = ( | L g | ϵ ' + | L f | ϵ ' + ϵ ' 2 ) | h | ,
(37)

whence

| θ ( h ) / h | < ( | L g | ϵ ' + | L f | ϵ ' + ϵ ' 2 ) < ϵ , | θ ( h ) / h | < ( | L g | ϵ ' + | L f | ϵ ' + ϵ ' 2 ) < ϵ ,
(38)

as desired.

Exercise 5

  1. Derive the familiar formulas for the derivatives of the elementary transcendental functions:
    exp'=exp,sin'=cos,,sinh'=cosh,cosh'=sinhandcos'=-sin.exp'=exp,sin'=cos,,sinh'=cosh,cosh'=sinhandcos'=-sin.
    (39)
  2. Define a function ff as follows. f(z)=cos2(z)+sin2(z).f(z)=cos2(z)+sin2(z). Use part (a) and the Chain Rule to show that f'(z)=0f'(z)=0 for all z∈C.z∈C. Does this imply that cos2(z)+sin2(z)=1cos2(z)+sin2(z)=1 for all complex numbers z?z?
  3. Suppose ff is expandable in a Taylor series around the point c:c:f(z)=∑n=0∞an(z-c)nf(z)=∑n=0∞an(z-c)n for all z∈Br(c).z∈Br(c). Prove that ff is differentiable at each point of the open disk Br(c),Br(c), and show that
    f'(z)=∑n=1∞nan(z-c)n-1.f'(z)=∑n=1∞nan(z-c)n-1.
    (40)
    HINT: Use Theorem 5 and the chain rule.

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