Obviously there need to be simple ways of finding the derivative when it exists. There are
rules of derivatives which make finding the derivative very easy. In the following theorem,
the derivative could refer to right or left derivatives as well as regular derivatives.
Theorem 5.2.1Let a,b be numbers and suppose f^{′}
(t)
and g^{′}
(t)
exist. Then thefollowing formulas are obtained.
′ ′ ′
(af + bg) (t) = af (t)+ bg(t). (5.1)
(5.1)
′ ′ ′
(fg) (t) = f (t)g(t) + f (t) g(t). (5.2)
(5.2)
The formula, 5.2is referred to as the product rule.
If f^{′}
(g(t))
exists and g^{′}
(t)
exists, then
(f ∘g)
^{′}
(t)
also exists and
′ ′ ′
(f ∘g) (t) = f (g (t))g (t).
This is called the chain rule. In this rule, for the sake of simiplicity, assume thederivatives are real derivatives, not derivatives from the right or the left.If f
(t)
= t^{n}where n is any integer, then
f′(t) = ntn−1. (5.3)
(5.3)
Also, whenever f^{′}
(t)
exists,
′ f (t+-h)−-f-(t)
f (t) = lhim→0 h
where this definition can be adjusted in the case where the derivative is a right or leftderivative by letting h > 0 or h < 0 only and considering a one sided limit.This isequivalent to
f′(t) = lim f-(s)-− f-(t)
s→t t− s
with the limit being one sided in the case of a left or right derivative.
= 0 and so if this difference is not equal to 0, it will be very small,
provided h is small enough, and the difference quotient is close to f^{′}
(g(t))
. The other
case is that g
(t+ h)
= g
(t)
. In this case, H
(h)
= f^{′}
(g (t))
which is as close as it is
possible to be to f^{′}
(g (t))
. Then
f (g(t+-h))−-f (g(t))-= H (h) g(t+-h)−-g-(t)
h h
and so,
lim f-(g(t+-h))-− f-(g(t)) = lim H (h) g-(t+-h)-−-g(t) = f′(g (t))g′(t).
h→0 h h→0 h
The last claim follows from Example 5.1.2 in case n is a positive integer. If n is 0,
then the claim is obvious. It remains to consider the case where n is a negative integer.
First consider f
(t)
= t^{−1}. Then
f (t+ h)− f (t) t+1h-− 1t 1
------h-------= ---h--- = −(t+-h)t
For all t≠0, the limit of this last expression is −
-1
t2
using the properties of the limit.
Therefore, if f
(t)
= t^{−n}, then
n −1
f (t) = (t )
and so, by the chain rule and what was just shown for positive exponent n,
f^{′}
(t)
=
(− 1)
(tn)
^{−2}nt^{n−1} = −nt^{−(n+1)
} showing that the claim holds in this case also.
■
Corollary 5.2.2Let f^{′}
(t)
,g^{′}
(t)
bothexist and g
(t)
≠0, then the quotient ruleholds.
( ) ′ ′ ′
f- = f-(t)g-(t)−-f2-(t)g-(t)-
g g(t)
Proof: This is left to you. Use the chain rule and the product rule. ■
Higher order derivatives are defined in the obvious way.
′′ ′ ′
f ≡ (f )
etc. Also the Leibniz notation is defined by
dy-= f′(x) where y = f (x)
dx
and the second derivative is denoted as
2
d-y
dx2
with various other higher order derivatives defined similarly.
The chain rule has a particularly attractive form in Leibniz’s notation. Suppose
y = g
(u)
and u = f
(x)
. Thus y = g ∘ f
(x )
. Then from the above theorem
(g∘ f)
^{′}
(x)
= g^{′}
(f (x))
f^{′}
(x)
= g^{′}
(u)
f^{′}
(x)
or in other words,
dy-= dy-du.
dx du dx
Notice how the du cancels. This particular form is a very useful crutch and is used
extensively in applications.