Lecture 19: Differentiability
Again we look at how to build differentiable functions from others.
Proposition 0.1. Let f, g : (a, b) → R be differentiable at x. Then the following functions
are differentiable with derivatives:
1. (f + g)0 (x) = f 0 (x) + g 0 (x)
2. (f g)0 (x) = f 0 (x)g(x) + f (x)g 0 (x).
3. (f /g)0 (x) =
f 0 (x)g(x)−f (x)g 0 (x)
g 2 (x)
if g(x) 6= 0.
Proof. For the first we just use properties of limits:
(f + g)(y) − (f + g)(x)
f (y) − f (x)
g(y) − g(x)
= lim
+ lim
= f 0 (x) + g 0 (x) .
y→x
y→x
y→x
y−x
y−x
y−x
lim
For the second, we write
(f g)(y) − (f g)(x) = (f (y) − f (x))g(y) + f (x)(g(y) − g(x)) ,
divide by y − x and take a limit:
(f g)(y) − (f g)(x)
f (y) − f (x)
g(y) − g(x)
= lim
lim g(y) + f (x) lim
.
y→x
y→x
y→x
y→x
y−x
y−x
y−x
lim
As g is differentiable at x, it is also continuous, so g(y) → g(x) as x → y. This gives the
formula.
The last property can be derived in a similar fashion:
1
[f (y)g(x) − f (x)g(y)]
g(y)g(x)
1
=
[g(x)(f (y) − f (x)) − f (x)(g(y) − g(x))] .
g(y)g(x)
(f /g)(y) − (f /g)(x) =
Dividing by y − x and taking the limit gives the result.
Again, from this proposition, we find that all polynomials are differentiable everywhere
as are rational functions wherever the denominator is nonzero. The next way to build
differentiable functions is to compose:
Theorem 0.2 (Chain rule). Let f : (a, b) → (c, d) be differentiable at x0 and g : (c, d) → R
be differentiable at f (x0 ). Then g ◦ f is differentiable at x0 with derivative
(g ◦ f )0 (x0 ) = f 0 (x0 )g 0 (f (x0 )) .
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Proof. We will want to use a division by f (y) − f (x0 ) for y 6= x0 , so we must first deal with
the case that this could be 0. If there exists a sequence (xn ) in (a, b) with xn → x0 but
xn 6= x0 for all n with f (xn ) = f (x0 ) for infinitely many n, we would have
f (y) − f (x0 )
f (xn ) − f (x0 )
= lim
=0,
y→x
n→∞
y − x0
xn − x 0
f 0 (x0 ) = lim
so the right side of the equation in the theorem would be 0. The left side would also be zero
for a similar reason:
lim
y→x0
(g ◦ f )(y) − (g ◦ f )(x0 )
g(f (xn )) − g(f (x0 ))
= lim
=0.
n→∞
y − x0
xn − x0
In the other case, every sequence (xn ) in (a, b) with xn → x0 and xn 6= x0 has f (xn ) =
f (x0 ) for at most finitely many n. Then as f is continuous at x0 , we have f (xn ) → f (x0 )
with an 6= f (x0 ) for all n and so
g(f (xn )) − g(f (x0 ))
(g ◦ f )(y) − (g ◦ f )(x)
= lim
n→∞
y→x
y−x
xn − x 0
g(f (xn )) − g(f (x0 ))
f (xn ) − f (x0 )
= lim
· lim
n→∞
n→∞
f (xn ) − f (x0 )
xn − x0
0
0
= g (f (x0 ))f (x0 ) .
lim
Examples.
1. We know f (x) = |x| is continuous but not differentiable. To go one level deeper,
consider
(
x2
x≥0
.
f (x) =
2
−x x < 0
The derivative at 0 is
f (0 + h)
=0,
h→0
h
lim
and the derivative elsewhere is
(
2x
x>0
f 0 (x) =
.
−2x x < 0
Note that f 0 is continuous. Then we say f ∈ C 1 (or f is in class C 1 ). However the
second derivative does not exist.
2. The function
(
x3
f (x) =
−x3
x≥0
x<0
is in class C 2 , as it has two continuous derivatives. But it is not three times differentiable.
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3. Generally, the function
(
xn
f (x) =
−xn
x≥0
, n≥1
x<0
is in class C n−1 , meaning that it has n − 1 continuous derivatives. But it is not n times
differentiable.
Mean value theorem
We begin by looking at local extrema.
Definition 0.3. For X a metric space, let f : X → R. We say that x0 ∈ X is a local
maximum for f if there exists r > 0 such that for all x ∈ Br (x0 ) we have f (x) ≤ f (x0 ).
Similarly x0 is a local minimum for f if there exists r > 0 such that for all x ∈ Br (x0 ) we
have f (x) ≥ f (x0 ).
In the case that X is R, if f is differentiable at a local extreme point, then the derivative
must be zero.
Proposition 0.4. Let f : (a, b) → R and suppose that c ∈ (a, b) is a local extreme point for
f . If f 0 (c) exists then f 0 (c) = 0.
Proof. Let c be a local max such that f 0 (c) exists. Then there exists r > 0 such that for all
y with |y − c| < r, we have f (y) ≤ f (c). Therefore, looking at only right limits,
lim+
f (y) − f (c)
≤0.
y−c
lim−
f (y) − f (c)
≥0.
y−c
y→c
Looking only at left limits,
y→c
0
Putting these together, we find f (c) = 0. The argument for local min is similar.
Theorem 0.5 (Rolle’s theorem). For a < b, let f : [a, b] → R be continuous such that f is
differentiable on (a, b). If f (a) = f (b) then there exists c ∈ (a, b) such that f 0 (c) = 0.
Proof. If f is constant on the interval then clearly the statement holds. Otherwise for some
d ∈ (a, b) we have f (d) > f (a) or f (d) < f (a). Let us consider the first case; the second is
similar. By the extreme value theorem, f takes a maximum on [a, b] and since f (d) > f (a)
this max cannot occur at a or b. So it occurs at some c ∈ (a, b). Then c is a local max as
well, so we can apply the previous proposition to find f 0 (c) = 0.
An important corollary is the following.
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Corollary 0.6 (Mean value theorem). For a < b let f : [a, b] → R be continuous such that
f is differentiable on (a, b). There exists c ∈ (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Proof. Define L(x) to be the line that connects the points (a, f (a)) and (b, f (b)):
L(x) =
f (b) − f (a)
(x − a) + f (a) .
b−a
Then the function g = f − L satisfies g(a) = g(b) = 0. It is also continuous on [a, b]
and differentiable on (a, b). Therefore by Rolle’s theorem, we can find c ∈ (a, b) such that
g 0 (c) = 0. This gives
0 = g 0 (c) = f 0 (c) − L0 (c) = f 0 (c) −
f (b) − f (a)
,
b−a
implying the corollary.
The mean value theorem has a lot of consequences. It is one of the central tools to
analyze derivatives.
Corollary 0.7. Let f : (a, b) → R be differentiable.
1. If f 0 (x) ≥ 0 for all x ∈ (a, b) then f is non-decreasing.
2. If f 0 (x) ≤ 0 for all x ∈ (a, b) then f is non-increasing.
3. If f 0 (x) = 0 for all x ∈ (a, b) then f is constant.
Proof. Suppose first that f 0 (x) ≥ 0 for all x ∈ (a, b). To show f is non-decreasing, let c < d
in (a, b). By the mean value theorem, there exists x0 ∈ (c, d) such that
f 0 (x0 ) =
f (d) − f (c)
.
d−c
But this quantity is nonnegative, giving f (d) ≥ f (c). The second follows by considering −f
instead of f . The third follows from the previous two.
Example: Power series. We will derive some results about power series because
they will
P
n
help us on the problem set to define trigonometric functions. Let f (x) = ∞
a
n=0 n x be a
power series with radius of convergence R > 0. We wish to show that
• f is differentiable on (−R, R).
P
n−1
• The power series ∞
also has radius of convergence R.
n=0 nan x
P
n−1
• For all x ∈ (−R, R), f 0 (x) = ∞
.
n=0 nan x
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Step 1. The power series
need a lemma.
P∞
n=0
nan xn−1 also has radius of convergence R. To show this, we
Lemma 0.8. Suppose that (xn ) and (yn ) are non-negative real sequences such that xn →
x > 0. Then
lim sup xn yn = x lim sup yn .
n→∞
n→∞
Proof. We will use the definition from the homework that lim supn→∞ bn is the supremum
of all subsequential limits of (bn ). Let S be the set of subsequential limits of (yn ) and T the
corresponding set for (xn yn ). We will prove the case that S and T are bounded above; the
other case is left as an exercise.
We claim that
xS = T , where xS = {xs : s ∈ S} .
To prove this, let a ∈ xS. Then there exists a subsequence (ynk ) such that ynk → a/x.
Now xnk ynk → xa/x = a, giving that a ∈ T . Conversely, let b ∈ T so that there exists a
subsequence (xnk ynk ) such that xnk ynk → b. Then ynk = xnk ynk /xnk → b/x. This means
that b = xb/x ∈ xS.
To finish the proof we show that sup T = x sup S. First if t ∈ T we have t/x ∈ S, so
t/x ≤ sup S. Therefore t ≤ x sup S and sup T ≤ x sup S. Conversely if s ∈ S then xs ∈ T ,
so xs ≤ sup T , giving s ≤ (1/x) sup T . This means sup S ≤ (1/x) sup T and therefore
sup T ≥ x sup S.
P
n−1
, we use the root test:
To find the radius of convergence of ∞
n=0 nan x
lim sup (n|an |)1/n = lim sup n1/n |an |1/n .
n→∞
n→∞
Since n1/n → 1 weP
can use the previous lemma to get a limsup of 1/R, where R is the radius
n
of convergence of ∞
n=0 an x . This means the radius of convergence of the new series is also
R.
P
n
Step 2. The function f given by f (x) = ∞
n=0 an x is differentiable at x = 0.
To prove this, we use 0 < |x| < R/2 and compute
P∞
∞
an xn − a0 X
f (x) − f (0)
= n=0
=
an xn−1 .
x−0
x
n=1
Pulling off the first term,
X
∞
X
f (x) − f (0)
∞
n−1 n−2 − a1 = an x = |x| an x .
x−0
n=2
n=2
We can use the triangle inequality for the last sum to get
∞
∞
X
X
f (x) − f (0)
n−2
−
a
≤
|x|
|a
||x|
≤
|x|
|an |(R/2)n−2 .
1
n
x−0
n=2
n=2
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By the ratio test, the last series converges, so setting C equal to it, we find
f (x) − f (0)
≤ C|x| .
−
a
1
x−0
Now we can take the limit as x → 0 and find
f (x) − f (0)
f (x) − f (0)
lim − a1 = 0 , or lim
= a1 .
x→0
x→0
x−0
x−0
This means f 0 (0) = a1 .
Step 3 is next time.
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