Math 313: Lecture 25
M. K. Vemuri
Last homework
It is expected that you have already read articles 4.20 (p91–92),
4.29 (proof) (p96), 4.31–4.34 (p97–98).
The derivative of a real function
Definition
Let f : [a, b] → R. For any x ∈ [a, b] let
f (t) − f (x)
φx (t) = (a < t < b, t ̸= x),
t −x
and define
f ′ (x) = lim φx (t),
t→x
provided this limit exists.
The function f ′ , whose domain is the set of points at which the
above limit exists, is called the derivative of f . If f ′ (x) is defined,
we say f is differentiable at x. If f ′ (x) exists for all x ∈ E , we say
that f is differentiable on E .
The derivative of a real function (contd)
Theorem
If f is differentiable at x then f is continuous at x.
Proof.
lim f (t) = lim [f (x) + φx (t)(t − x)]
t→x t→x
= lim f (x) + lim φx (t) lim (t − x)
t→x t→x t→x
= f (x) + f ′ (x)(0) = f (x).
The derivative of a real function (contd)
Theorem
Suppose f and g are differentiable at x. Then f + g and fg are
differentiable at x and
(a). (f + g )′ (x) = f ′ (x) + g ′ (x);
(b). (fg )′ (x) = f ′ (x)g (x) + f (x)g ′ (x).
Furthermore, if g (x) ̸= 0, then f /g is differentiable at x and
g (x)f ′ (x)−g ′ (x)f (x)
(c). (f /g )′ (x) = (g (x))2
.
The derivative of a real function (contd)
Proof.
We will prove (b). The other parts are left to the reader.
(fg )(t) − (fg )(x) f (t)[g (t) − g (x)] + [f (t) − f (x)]g (x)
lim = lim
t→x t −x t→x t −x
g (t) − g (x)
= lim f (t) lim
t→x t→x t −x
f (t) − f (x)
+ lim g (x)
t→x t −x
= f (x)g ′ (x) + f ′ (x)g (x).
Example
If n ∈ W, then the function x n is differentiable everywhere and
f ′ (x) = nx n−1 . Therefore, polynomials are differentiable
everywhere, and rational functions are differentiable where they are
defined.
The derivative of a real function (contd)
Theorem (Chain rule)
Suppose f is continuous on [a, b] and differentiable at x, and g is
defined on an interval I containing the range of f and is
differentiable at f (x). Then g ◦ f is differentiable at x and
(g ◦ f )′ (x) = g ′ (f (x))f ′ (x).
Proof. Define
( g (s)−g (f (x))
s−f (x) , s ̸= f (x),
ψ(s) =
g ′ (f (x)), s = f (x).
Then ψ is continuous at f (x).
The derivative of a real function (contd)
Therefore
g (f (t)) − g (f (x))
(g ◦ f )′ (x) = lim
t→x t −x
f (t) − f (x)
= lim ψ(f (t))
t→x t −x
= ψ(f (x))f (x) = g ′ (f (x))f ′ (x).
′
Mean Value Theorems
Definition
Let f be a real valued function defined on a metric space X . We
say that f has a local maximum at a point p ∈ X if there exists
r > 0 such that f (q) ≤ f (p) for all q ∈ Nr (p).
Theorem
Let f : [a, b] → R. If f has a local maximum at a point x ∈ (a, b)
and f ′ (x) exists, then f ′ (x) = 0.
Proof. By hypothesis there exists r > 0 such that
|t − x| < r =⇒ f (t) ≤ f (x). If x − r < t < x then
f (t) − f (x)
≥ 0,
t −x
so f ′ (x) ≥ 0.
Mean Value Theorems (contd)
If x < t < x + r then
f (t) − f (x)
≤ 0,
t −x
so f ′ (x) ≤ 0.
Theorem (Mean Value Theorem)
If f : [a, b] → R is continuous and f is differentiable on (a, b) then
there exists x ∈ (a, b) such that
f (b) − f (a) = (b − a)f ′ (x).
Proof. Let
f (b) − f (a)
h(x) = f (x) − (x − a).
b−a
We claim that there exists x ∈ (a, b) such that h′ (x) = 0.
Mean Value Theorems (contd)
Since [a, b] is compact, there exist x1 , x2 ∈ [a, b] such that
h(x1 ) ≤ h(x) ≤ h(x2 ) for all x ∈ [a, b]. Observe that
h(a) = f (a) = h(b). Therefore if {x1 , x2 } ⊆ {a, b}, then h is
constant, and we may take (say) x = (a + b)/2. If x1 ∈ (a, b),
then −h has a local maximum at x1 , and so h′ (x1 ) = 0, in which
case we take x = x1 . If x2 ∈ (a, b), then h has a local maximum at
x2 , and so h′ (x2 ) = 0, in which case we take x = x2 .
Since h′ (x) = 0, we have f ′ (x) = f (b)−f
b−a
(a)
as required.
Sometimes the following statement (see book for proof) is needed:
Theorem (Generalized Mean Value Theorem)
If f , g : [a, b] → R are continuous and are differentiable on (a, b)
then there exists x ∈ (a, b) such that
[f (b) − f (a)]g ′ (x) = [g (b) − g (a)]f ′ (x).
