MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
6. Differentiation
f (a + h) − f (a)
Definition. The function f is differentiable at a if
exists.
h
0
In this case the limit is denoted by f (a) and is called the derivative if f at a.
Example. Clearly f (x) = x2 is differentiable everywhere on R.
(a + h)2 − a2
2ah + h2
(1) Let a ∈ R.
=
= 2a + h → 2a as h → 0.
h
h
(2) f (x) = |x| is continuous everywhere and is differentiable if x 6= 0. If a = 0 lim
h→0
f (0 + h) − f (0)
=
h
|h|
|h|
h
|h| − 0
= lim
. If h ≥ 0 lim
= lim
= lim 1 = 1 (right hand derivative) If h < 0
h→0
h→0 h
h→0+ h
h→0+ h
h→0
h
|h|
−h
lim
= lim
= lim −1 = −1 (left hand derivative) Hence f is not differentiable at
h→0− h
h→0− h
h→0
a = 0.
lim
(3)

1

x sin
f (x) =
x
0
x 6= 0
x=0
Recall that lim f (x) = 0 = f (0) and therefore f is continuous everywhere on R.
x→0
h sin h1 − 0
1
f (0 + h) − f (0)
= lim
= lim sin which does not exist. Hence
h→0
h→0
h
h
h
f (x) is not differentiable at 0.
But when a = 0,
(4)


f (x) =
x2 sin
0
1
x
x 6= 0
x=0
f is continuous at 0 (in fact everywhere) and f is differentiable at x = 0 (in fact everywhere).
h sin h1 − 0
1
= lim h sin = 0.
h→0
h→0
h
h
Theorem 6.1. Let f be differentiable at a then f is continuous at a.
lim
Proof. It suffice to prove lim f (a + h) = f (a).
h→0
f (a + h) − f (a)
· h}
h
f (a + h) − f (a)
= lim {
} lim h
h→0
h→0
h
0
= f (a) · 0 = 0.
lim {f (a + h) − f (a)} = lim {
h→0
h→0
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Theorem 6.2. Let f (x) and g(x) be differentiable at a and c a fixed constant. Then
(1) f + g is differentiable at a and
¯
¯
= f 0 (a) + g 0 (a),
f 0 (x) + g 0 (x)¯
x=a
1
2
¯
¯
(2) cf 0 (x)¯
x=a
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
= c0 f (a).
Theorem 6.3. Let f (x) and g(x) be differentiable at a then f (x)g(x) is differentiable at a and
¯
¯
(f (x)g(x))0 ¯
= f 0 (a)g(a) + f (a)g 0 (a).
x=a
Proof.
f (a + h)g(a + h) − f (a)f (a)
h
f (a + h)g(a + h) − f (a)g(a + h) + f (a)g(a + h) − f (a)f (a)
= lim
h→0
h
f (a + h) − f (a)
g(a + h) − f (a)
= lim g(a + h)(
) + lim f (a)(
)
h→0
h→0
h
h
f (a + h) − f (a)
g(a + h) − f (a)
= lim g(a + h) lim (
) + f (a) lim (
)
h→0
h→0
h→0
h
h
= g(a)f 0 (a) + f (a)g 0 (a).
lim
h→0
We used the fact that g is continuous at a.
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Theorem 6.4. If f and g are differentiable at a and g(a) 6= 0 then
f 0 ¯¯
g 0 (a)f (a) − g(a)f 0 (a)
( )¯
=
.
g x=a
(g(a))2
f
is differentiable at a and
g
Proof. We prove the special case when f ≡ 1. Since g is differentiable at a and so continuous at a.
Since also g(a) 6= 0 and we can find a δ > 0 such that g(a + h) 6= 0 for |h| < δ.
1
1
1
1 g(a) − g(a + h)
(
−
) = lim
|h| < δ
h→0 h g(a + h)
h→0 h g(a + h)(g(a))
g(a)
−1
g(a + h) − g(a)
= lim
lim
h→0 g(a + h)g(a) h→0
h
−1 0
=
g (a) by the continuity of g at a.
g(a)2
lim
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Theorem 6.5. (Chain Rule)
Let g be differentiable at a and let f be differentiable at g(a). Then f (g(a)) is differentiable when
x = a and
¯
¯
(f (g(x)))¯
= f 0 (g(a))g 0 (a).
x=a
Proof. Define

 f (g(a + h)) − f (g(a))
g(a + h) − g(a)
φ(h) =
 0
f (g(a))
if g(a + h) − g(a) 6= 0
.
if g(a + h) − g(a) = 0
We aim to prove φ(h) is continuous at h = 0 and limh→0 φ(h) = f 0 (g(a)). Let k = g(a + h) − g(a) 6= 0.
Since f is differentiable at g(a), hence
f (g(a) + k) − f (g(a))
= f 0 (g(a)).
k→0
k
That is, given ² > 0, we can find a δ 0 > 0 such that
¯
¯
¯ f (g(a) + k) − f (g(a))
¯
¯
¯
0
− f (g(a))¯ < ² whenever 0 < |k| < δ 0 .
¯
¯
¯
k
lim
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
3
Since g(x) is differentiable at a and hence continuous there.
We can find a δ > 0 such that
|g(a + h) − g(a)| = |k| < δ 0 whenever |h| < δ.
Hence
|φ(h) − f 0 (g(a))| < ² whenever k 6= 0 and |h| < δ.
But even if k = 0, (1) is still valid. Hence φ(h) is continuous at h = 0. We can now complete the proof.
For h 6= 0
g(a + h) − g(a)
f (g(a + h)) − f (g(a))
= φ(h) ·
h
h
even if g(a + h) = g(a). Therefore
f (g(a + h)) − f (g(a))
h
g(a + h) − g(a)
= lim φ(h) · lim
h→0
h→0
h
0
0
= f (g(a)) · g (a).
(f ◦ g)0 (a) = lim
h→0
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Definition. Let f be a function defined on a set of numbers A. A point x ∈ A is a maximum point
for f on A, is
f (y) ≤ f (x) for every y ∈ A.
Theorem 6.6. Let f be a function defined on (a, b). If x is a maximum (respectively minimum) point
for f on (a, b), and f is differentiable at x, then f 0 (x) = 0.
Proof. We choose h such that x + h is in (a, b), then
f (x) ≥ f (x + h)
since f (x) is a maximum in (a, b). Hence
f (x + h) − f (x) ≤ 0.
If h > 0
Hence
f (x + h) − f (x)
≤ 0.
h
f (x + h) − f (x)
≤ 0.
h→0+
h
lim
If h < 0,
f (x + h) − f (x)
≥ 0.
h
Hence
f (x + h) − f (x)
≥ 0.
h→0−
h
lim
Hence
lim
h→0−
That is, f 0 (x) = 0.
f (x + h) − f (x)
≥ 0.
h
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We have obtained results about f 0 by giving information about f . If we are given f 0 (x) = 0 we
cannot predict the general picture of f . For example if f 0 (x) = 0 for all x, can we deduce f is a
constant from it?
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MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
Theorem 6.7. Let f be differentiable on (a, b) and suppose a < s < t < b. Let ρ ∈ R such that
f 0 (s) < ρ < f 0 (t).
Then there is c ∈ (s, t) such that f 0 (c) = ρ.
Proof. Define a function G(x) = f (x) − ρx.
Then
G0 (s) = f 0 (s) − ρ < 0
G0 (t) = f 0 (t) − ρ > 0
Since G is continuous on [s, t] and hence attains a minimum at [s, t]. But G0 (s) > 0 and G0 (t) < 0
hence c 6= s or t and so c ∈ (s, t) and G0 (c) = 0. That is, f 0 (c) = ρ.
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Remark. We say that a g(x) has intermediate value property on I if for all s, t ∈ I and for any ρ
g(s) < ρ < g(t)
there is c ∈ (s, t) such that g(c) = ρ.
So if f is differentiable on I then f 0 has the intermediate value property. Any continuous function
also has this property.

1

x 6= 0 on [0, 1]
sin
However if f has this IMV property does not imply f is continuous. Consider
.
x
0
x=0
Theorem 6.8. (Rolle’s Theorem)
If f is continuous on [a, b] and differentiable on (a, b), and f (a) = f (b), then there is a number
ξ ∈ (a, b) such that f 0 (ξ) = 0.
Proof. Since f is continuous on [a, b], we deduce f attains its maximum or minimum in [a, b]. Suppose
f (x) is a maximum and x ∈ (a, b). Then by theorem 5.6 f 0 (x) = 0. If the maximum or minimum
occurs at the end point a or b, then f is constant. f 0 (x) = 0 for each x ∈ (a, b).
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Theorem 6.9. (Mean Value Theorem)
Let f be continuous on [a, b] and differentiable on (a, b), then there exists a ξ in (a, b) such that
f 0 (ξ) =
Proof. Consider the function g defined by
g(x) = f (x) −
Ã
f (b) − f (a)
.
b−a
!
f (b) − f (a)
(x − a) + f (a)
b−a
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
5
Then g(a) = 0 and g(b) = 0. Clearly g is continuous on [a, b] and differentiable in (a, b). Hence by
Rolle’s theorem, we can find a ξ in (a, b) such that g 0 (ξ) = 0. That is,
f 0 (ξ) =
f (b) − f (a)
.
b−a
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Corollary 6.10. If f is defined on the interval and f 0 (x) = 0 for all x in the interval, then f is
constant on the interval.
Proof. Let a, b be any two points in the interval with b > 0, then there is some x ∈ (a, b) such that
0 = f 0 (x) =
f (b) − f (a)
.
b−a
Hence f (b) = f (a). Since this is true for all a and b, f must be a constant.
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Theorem 6.11. If f and g are continuous on [a, b] and differentiable on (a, b), then there is a ξ in
(a, b) such that
(f (b) − f (a))g 0 (ξ) = (g(b) − g(a))f 0 (ξ).
Proof. Consider the function
G(x) = g(x)(f (b) − f (a)) − f (x)(g(b) − g(a)).
Then
G(a) = g(a)(f (b) − f (a)) − f (a)(g(b) − g(a))
= g(a)f (b) − f (a)g(b)
G(b) = g(b)(f (b) − f (a)) − f (b)(g(b) − g(a))
= −g(b)f (a)g (a)f (b) = G(a).
Clearly G(x) satisfies the criteria of the Rolle’s theorem. Hence there exists ξ ∈ (a, b) such that
G0 (ξ) = 0.
That is
(f (b) − f (a))g 0 (ξ) = (g(b) − g(a))f 0 (ξ).
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Theorem 6.12. (L’Hôpital’s rule)
f 0 (x)
f
exists. Then lim exists,
0
x→a g
x→a g (x)
Suppose that lim f (x) = 0 and lim g(x) = 0, and suppose that lim
x→a
x→a
and
f 0 (x)
f (x)
= lim
.
x→a g 0 (x)
x→a g(x)
lim
f 0 (x)
exists, implies f 0 , g 0 (x) exist in (a − δ, a + δ) except possibly
x→a g 0 (x)
when x = a and g 0 (x) 6= 0 in (a − δ, a + δ) except possibly x = a.
f and g may have removable discontinuity at x = 0 so we define f (a) = 0 = g(a). We claim that
there is an interval (a − δ, a + δ) such that g(x) 6= 0 in it. For suppose 0 < h < δ and g(a + h) = 0,
then the MVT implies 0 = g(a + h) − g(a) = h · g 0 (ah ) for some ah in (a, a + h) contradicting the fact
that g 0 6= 0 in that interval.
Proof. From the hypothesis lim
f (x) − f (a)
f 0 (ax )
f (x)
=
= 0
g(x)
g(x) − g(a)
g (ax )
ax ∈ (a, x) (or (x, a))
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MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
f 0 (x)
f 0 (x)
=
l.
So
given
²
>
0,
we
can
find
a
δ
>
0
such
that
|
−l| < ² whenever 0 < |x−a| < δ.
x→a g 0 (x)
g 0 (x)
Hence
f (x)
f (x) f 0 (ax )
f 0 (ax )
− l| ≤ |
− 0
|+| 0
− l|
|
g(x)
g(x)
g (ax )
g (ax )
< 0 + ² whenever 0 < |x − a| < δ.
Let lim
That is,
f (x)
f 0 (x)
= l = lim 0
.
x→a g (x)
x→a g(x)
lim
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The rule holds when a = ∞ and/or when f (x) → ∞ and g(x) → ∞ as x → a.
xn
xn−1
1
=
n
lim
= · · · = n! lim x = 0.
x
x
x→∞ e
n→∞ e
x→∞ e
Example. If n > 0, lim xn e−x = lim
x→∞
Inverse functions.
Definition. A function f is one-to-one if f (a) 6= f (b) whenever a 6= b. The inverse of f , if it exists,
denoted by f −1 , is the set of all pairs (a, b) for which (b, a) ∈ f .
See Chapter 4. Clearly f −1 exists if and only if f is one-to-one.
Theorem 6.13. If f is continuous and one-to-one on an interval, then f is either increasing or
decreasing on that interval.
Proof. Suppose a0 < b0 and we have either
f (b0 ) > f (a0 ) or f (a0 ) > f (b0 ).
So we may assume the first case. That is, f (b0 ) > f (a0 ). Let a1 and b1 be any two points in the
interval such that a1 < b1 and shall show the same inequality still hold.
Claim: if a0 and a1 in interval then
xt = (1 − t)a0 + ta1
0≤t≤1
also belongs to the interval.
Similarly for yt = (1 − t)b0 + tb1 , 0 ≤ t ≤ 1 (exercise).
Since a0 < b0 and a1 < b1 we have xt < y t for 0 ≤ t ≤ 1.
Define g(t) = f (yt ) − f (xt ), 0 ≤ t ≤ 1.
Clearly g is continuous on [0, 1] and not equals 0 there. Hence it must be either > 0 or < 0 on the
whole of [0, 1]. However g(0) = f (b0 ) − f (a0 ) > 0. Hence 0 < g(1) = f (b1 ) − f (a1 ).
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Theorem 6.14. If f is continuous and monotone on an interval, then f 0 is also continuous.
Proof. By the last theorem, we may assume that f is increasing. Suppose f (a) = b or a = f −1 (b). We
want to prove
Given ² > 0, we can find δ such that
|x − b| < δ =⇒ |f −1 (x) − f −1 (b)| < ²
(1)
or
− δ < x − f (z) < δ =⇒ a − ² < f −1 (x) < a + ²
or
f (a) − δ < x < δ + f (a) =⇒ a − ² < f −1 (x) < a + ²
Given ² > 0, we have
a − ² < a < a + ².
Hence
f (a − ²) < f (a) < f (a + ²).
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
7
Let δ = min{f (a) − f (a − ²), f (a + ²) − f (a)}. Then
f (a − ²) ≤ f (a) − δ < f (a) < f (a) + δ ≤ f (a + ²).
Consequently if f (a) − δ < x < δ + f (a)
f −1 (f (a − ²)) < f −1 (f (a) − δ) < f −1 (x) < f −1 (δ + f (a)) < f −1 (f (a + ²)),
That is,
a − ² < f −1 (x) < a + ²
and this is precisely (1).
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Theorem 6.15. If f is a continuous one-to-one function defined on I and f 0 (f −1 (a)) = 0, then f −1
is not differentiable at a.
Proof. (f (f −1 (a)))0 = f 0 (f −1 (a))(f −1 (a))0 = 1
Example. f (x) = x3 .
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