The Product and Quotient Rules
LRT
02/10/2017
As you have already guessed from the title, today we want
to discuss the rules for differentiating a product or a
quotient in terms of the functions which make up the
product or quotient and their derivatives.
Before going into the details of these formulas, we need to
pause for a general remark which explains why you will see
so little mention of continuity hereafter.
Differentiable functions are continuous
If a function f is differentiable at a
then it is continuous at a.
To show a function is continuous at a we need to see
lim f (x) = f (a).
x→a
An idea which is useful in many contexts when you need to
show two numbers are equal, see if you can show their
difference is 0.
In this case,
lim f (x) − f (a) = lim f (x) − lim f (a) = lim f (x) − f (a) .
x→a
x→a
x→a
x→a
f (x) − f (a)
f (x) − f (a)
lim
(x − a) = lim
lim (x − a) =
x→a
x→a
x→a
x−a
x−a
f 0 (a) · 0 = 0.
Given any function defined by a single formula, you will
soon be able to write down a formula for its derivative.
This shows it is differentiable and hence continuous.
The Product Rule
Let h(x) = f (x) · g(x). Then
f (x + h) · g(x + h) − f (x) · g(x)
h→0
h
h0 (x) = lim
What to do with this?
f (x + h) · g(x + h) − f (x) · g(x) = f (x + h) ·g(x + h) − f (x) · g(x + h) + f (x) · g(x + h) −
f (x) · g(x) = f (x + h) − f (x) g(x + h) + f (x) g(x + h) − g(x) .
f (x + h) · g(x + h) − f (x) · g(x)
=
h
f (x + h) − f (x) g(x + h) + f (x) g(x + h) − g(x)
=
h
g(x + h) − g(x)
f (x + h) − f (x)
g(x + h) + f (x)
h
h
Apply lim .
h→0
f (x + h) − f (x)
lim
lim g(x + h)+
h→0
h→0
h
g(x + h) − g(x)
lim f (x) lim
h→0
h→0
h
h0 (x) = f 0 (x)g(x) + f (x)g 0 (x)
The Product Rule
In English
The derivative of a product is
(the derivative of the first factor) times (the second factor)
plus
(the first factor) times (the derivative of the second factor).
Examples
d x2
:
dx
It is 2x by the Power Rule.
It is
dx
dx
x+x
= 1 · x + x · 1 = 2x by the Product Rule.
dx
dx
d f (x)2 0
: f (x)f (x) + f (x)f 0 (x) = 2f (x)f 0 (x) by the
dx
Product Rule.
More examples later.
The Quotient Rule
Let h(x) =
f (x)
. Then
g(x)
0
h (x) = lim
h→0
f (x+h)
g(x+h)
−
h
f (x)
g(x)
= lim
h→0
f (x+h)g(x)−f (x)g(x+h)
g(x+h)g(x)
h
Subtract and add f (x + h)g(x) and use your limit theorems
to get
f 0 (x)g(x) − f (x)g 0 (x)
h0 (x) =
2
g(x)
The Quotient Rule
In English
The derivative of a quotient is
(the derivative of the top factor) times (the bottom factor)
minus
(the top factor) times (the derivative of the bottom factor),
the whole thing divided by the bottom factor squared.
Examples
1
d g(x)
10 g(x) − 1 · g 0 x
0 · g(x) − g 0 x
d g(x)−1
=
=
=
=
dx
dx
g(x)2
g(x)2
g 0 (x)
−
= (−1)g(x)−2 g 0 (x).
g(x)2
h(t) =
t2
t
:
+1
h0 (t) =
1 · (t2 + 1) − t(2t)
t2 + 1 − 2t2
1 − t2
=
=
(t2 + 1)2
(t2 + 1)2
(t2 + 1)2
1
-5
-4
-3
-2
-1
0
-1
1
2
3
4
5
Here is a new sort of problem. Given a table of values and
values of the derivative, compute other values and
derivatives. Given this table of values
f
f0
g
g0
1 2
0 -1
-2 -1
2 0
2 -3
3 4 5 6
2 -1 3 4
1 3 2 -2
5 6 -1 2
11 7 -3 -1
0
f
f
find
(3) and
(3).
g
g
f
f (3)
2
(3) =
= :
g(3)
5
g 0
f 0 (3)g(3) − f (3)g 0 (3)
1 · 5 − 2 · 11
f
(3) =
=
=
g
g(3)2
52
5 − 22
17
=− .
25
25
Find (f · g)(2) and (f · g)0 (2).
f
f0
g
g0
1 2
0 -1
-2 -1
2 0
2 -3
3 4 5 6
2 -1 3 4
1 3 2 -2
5 6 -1 2
11 7 -3 -1
(f · g)(2) = (−1)(0) = 0:
(f · g)0 (2) = f 0 (2)g(2) + f (2)g 0 (2) = (−1) · 0 + (−1)(−3) = 3.
Revenue
Revenue for a company selling stuff can be computed as
R=p·x
where p is the unit price and x is the quantity sold.
Given a supply curve p = s(x) for the product, revenue is
now a function of a single variable x:
R(x) = s(x) · x
A question of some interest to management is how will our
revenue change if we change our production level.
To come up with a precise mathematical version of this
question/idea, we could ask for the instantaneous rate of
change in revenue with respect to production. In terms of
dR
formulas, what is
?
dx
d s(x) · x
dx
dR
=
= s0 (x) · x + s(x) ·
= s0 (x) · x + s(x)
dx
dx
dx
In a typical supply curve, s0 (x) < 0 so the question of
whether R is increasing or decreasing is a bit delicate. The
s(x) term is the current selling price which is positive for
any sensible business. The s0 (x) · x term is negative. If x is
large, even a small rate of change of the price will result in
revenue decreasing. Note s0 (x) is the instantaneous rate of
change in price with respect to production.
However, if p is large compared to x (think Ferrari) then
even a fairly hefty rate of change in price will still result in
an increase in revenue.
More Examples
Last time we computed the derivative of
x3 − 4x2 + 8
as f 0 (x) = 4x−2 − 24x−4 .
f (x) =
x3
We can also use the Quotient Rule:
f 0 (x) =
=
(3x2 − 8x)x3 − (x3 − 4x2 + 8)(3x2 )
(x3 )2
3x5 − 8x4 − (3x5 − 12x4 + 24x2 )
4x4 − 24x2
=
x6
x6
Since f (x) = (x3 − 4x2 + 8)x−3 we can also use the Product
Rule.
f 0 (x) = (3x2 − 8x)x−3 + (x3 − 4x2 + 8)(−3x−4 ) =
3x−1 − 8x−2 + (−3)x−1 − 4(−3)x−2 − 24x−4 =
4x−2 − 24x−4 .
1
The slope is to be − . At any point x, the slope is
2
1(x − 1) − (x + 1)(1)
−2
0
f (x) =
=
.
(x − 1)2
(x − 1)2
Hence we are looking for x such that
−2
1
=−
2
(x − 1)
2
Hence 4 = (x − 1)2 , x − 1 = ±2 so x = 3 and x = −1.
I could have jazzed up the problem a bit by asking for
points where the tangent line is parallel to x + 2y = 0 or for
points where the tangent line is perpendicular to
y = 2x + 4.
C 0 (t) =
−0.2t2 + 0.2
0.2(t2 + 1) − (0.2t)(2t)
=
(t2 + 1)2
(t2 + 1)2