Math 131.
Product and Quotient Rules, and Higher-Order Derivatives Larson Section 2.3
Let us observe first that
d
[f (x)g(x)] 6= f 0 (x)g 0 (x). Indeed, let f (x) = x and g(x) = x3 , then
dx
d
d
d 4
[f (x)g(x)] =
[x · x3 ] =
[x ] = 4x3
dx
dx
dx
but this is not equal to f 0 (x)g 0 (x) = 1 · 3x2 = 3x3 . The rule for the derivative of a product of
functions is more complicated than just multiplying their derivatives.
The Product Rule. If f and g are both differentiable at x, then so is f g and
d
[f (x)g(x)] = g(x)f 0 (x) + f (x)g 0 (x)
dx
Exercise. (a) Write the definition of derivative.
(b) Use the definition of derivative to prove the product rule.
Solution: (a) The derivative of f at x is
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
provided the limit exists.
(b)
f (x + h)g(x + h) − f (x)g(x)
d
(f (x)g(x)) = lim
h→0
dx
h
f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x − h) − f (x)g(x)
= lim
h→0
h
f (x + h) − f (x)
g(x + h) − g(x)
= lim g(x + h)
+ f (x)
h→0
h
h
0
0
= g(x)f (x) + f (x)g (x)
Example 1. Let f (x) = 24x(sin x + cos x). Find (a) f 0 (x) and (b) f 0
Solution: Given f (x) = 24x(sin x + cos x).
(a) First compute the derivative of f .
f 0 (x) = 24(sin x + cos x) + 24x(cos x − sin x)
= 24(1 − x) sin x + 24(1 + x) cos x.
π
6
(b) Using part (a), we find
π √ π
π
f0
= 12 1 −
+ 12 3 1 +
6
6
√6
√
= 12(1 + 3) + 2π( 3 − 1)
Similarly, there is a rule for the derivative of a quotient of two differentiable functions, and like
the product rule, it is not as simple as dividing the derivatives of the two functions: try this
f 0 (x)
f (x)
is not 0
. Thus we need:
with f (x) = x3 and g(x) = x to observe the derivative of
g(x)
g (x)
The Quotient Rule. Supose f and g are both differentiable at x, then so is
f
provided
g
g(x) 6= 0 and
d f (x)
g(x)f 0 (x) − f (x)g 0 (x)
=
dx g(x)
[g(x)]2
Exercise. Use the definition of derivative to prove the quotient rule.
Solution: The derivative of f at x is
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
provided the limit exists, and so
d f (x)
f (x + h)/g(x + h) − f (x)/g(x)
= lim
h→0
dx g(x)
h
f (x + h)g(x) − f (x)g(x + h)
= lim
h→0
hg(x)g(x + h)
f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h)
= lim
h→0
hg(x)g(x + h)
g(x)[f (x + h) − f (x)g(x)] f (x)[g(x + h) − g(x)]
= lim
−
h→0
hg(x)g(x + h)
hg(x)g(x + h)
g(x) · limh→0
f (x+h)−f (x)
h
− f (x) limh→0
h→0
limh→0 g(x)g(x + h)
0
0
g(x)f (x) − f (x)g (x)
=
[g(x)]2
= lim
g(x+h)−g(x)
h
Derivatives of the six trigonometric Functions:
d
[sin x] =
dx
cos x
d
[cos x] =
dx
− sin x
d
[tan x] =
dx
sec2 x
d
[cot x] =
dx
− csc2 x
d
[sec x] =
dx
sec x tan x
d
[csc x] =
dx
− csc x cot x
Look for patterns above, for example compare the derivatives of the function with the cofunctions, and decide how you can tell which derivatives have a negative sign.
Exercise. Use the quotient rule to find the derivatives of tan x, cot x, csc x and sec x by first
writing these functions in terms of sin x and cos x.
Solution: For the derivative of tan x, we compute
d
d sin x
[tan x] =
dx
dx cos x
(sin x)0 (cos x) − (sin x)(cos x)0
=
cos2 x
(cos x)(cos x) − (sin x)(− sin x)
=
cos2 x
2
2
1
cos x + sin x
=
= sec2 x
=
2
cos x
cos2 x
For the derivative of cot x, we compute
d
d h cos x i
[cot x] =
dx
dx sin x
(cos x)0 (sin x) − (cos x)(sin x)0
=
sin2 x
−(sin x)(sin x) − (cos x)(cos x)
=
sin2 x
cos2 x + sin2 x
1
= −
= − 2 = − csc2 x
2
sin x
sin x
For the derivative of csc x, we compute
d
1
d
[csc x] =
dx
dx sin x
(1)0 (sin x) − (1)(sin x)0
=
sin2 x
0 − (1)(cos x)
=
sin2 x
1
cos x
·
= − csc x cot x
= −
sin x sin x
For the derivative of sec x, we compute
d
d
1
[sec x] =
dx
dx cos x
(1)0 (cos x) − (1)(cos x)0
=
cos2 x
0 − (1)(− sin x)
=
cos2 x
1
sin x
=
·
= sec x tan x
cos x cos x
Two basic examples of using the quotient rule are as follows.
Example 2. Let g(x) =
Solution: g 0 (x) =
3x2 − 2
. Find g 0 (x).
2x + 1
6x(2x + 1) − (3x2 − 2)(2)
6x2 + 6x + 4
=
(2x + 1)2
(2x + 1)2
√
x−1
; find (a) f 0 (x), and (b) f 0 (7).
Example 3. Let f (x) = √
x+1
√
x−1
x1/2 − 1
Solution: (a) Given f (x) = √
and use the quotient rule
we write f (x) = 1/2
x +1
x+1
to find:
+ 1) − 12 x−1/2 (x1/2 − 1)
(x1/2 + 1)2
1
+ 12 x−1/2 − 12 + 21 x−1/2
1x−1/2
2
√
=
=
(x1/2 + 1)2
( x + 1)2
1
= √ √
x( x + 1)2
f 0 (x) =
1 −1/2 1/2
x
(x
2
1
and then (b) f 0 (7) = √ √
.
7( 7 + 1)2
Higher-order Derivatives. As we have seen, the derivative of a function is again a function,
and so it may well have a derivative, which will be called the second derivative. Then the
derivative of the second derivative will be called the third derivative . . . and so on.
Some notation of this is as follows:
First Derivative:
y0,
f 0 (x),
d
[f (x)],
dx
dy
dx
Second Derivative:
y 00 ,
f 00 (x),
d2
[f (x)],
dx2
d2 y
dx2
Third Derivative:
y 000 ,
f 000 (x),
d3
[f (x)],
dx3
d3 y
dx3
Fourth Derivative:
y (4) ,
f (4) (x),
d4
[f (x)],
dx4
d4 y
dx4
nth Derivative:
y (n) ,
f (n) (x),
dn
[f (x)],
dxn
dn y
dxn
Example 4. In Physics, we have that the derivative of position s(t) is velocity v(t) and the
derivative of velocity is acceleration a(t). Therefore,
v(t) = s0 (t)
and
a(t) = v 0 (t) = s00 (t)
On a planet in a distant galaxy an object is thrown upward at time t = 0. After t seconds its
height is y = −7.1t2 + 18.3t + 2.3 meters above the ground.
(a) From what height was the object thrown?
(b) What was the initial velocity of the object?
(c) What is the acceleration due to gravity on the planet?
Solution: (a) The initial height occurs at time t = 0, so the initial height of the object
was y(0) = 2.3 meters above the ground.
(b) The velocity, at time t, is given by y 0 = −14.2t + 18.3, and so the initial velocity was
y 0 (0) = 18.3 meters per second.
(c) The acceleration due to gravity is y 00 = −14.2 m/s2 .
A couple of examples involving higher order derivatives are as follows.
Example 5. (a) Let y = csc x find y 00 .
(b) Let y = cot x find y 00 .
Solution: (a) y 0 = − csc x cot x and so
y 00 = −[(csc x)0 (cot x) + (csc x)(cot x)0 ]
= −[− csc x cot x cot x − csc x csc2 x]
= csc x(cot2 x + csc2 x)
(b) y 0 = − csc2 x and so (without knowing the chain rule, but only the product rule):
y 00 = −[(csc x)0 (csc x) + (csc x)(csc x)0 ]
= −[−2 csc x csc x cot x]
= 2 csc2 x cot x
Example 6. If f (x) = 9 +
9
x
(a) f 0 (x)
(b) f 0 (5)
+
9
x4
find the following:
(c) f 00 (x)
(d) f 00 (5)
Solution: Frist, write f using negative exponents
f (x) = 9 + 9x−1 + 9x−4
The answers, unsimplified, are then
(a) f 0 (x) = −9x−2 − 36x−5
(b) f 0 (5) = − 592 −
(c) f 00 (x) = 18x−3 + 180x−6
(d) f 00 (5) =
18
53
+
36
55
180
56
Also, just because you have product and quotient rules, if you can avoid using them by simplifying the function first, it is often simpler to do so.
√
2 x
4
Example 7. For this problem, let f (x) = x πx + 15 − 1 . This is to emphasize to avoid
x
using the product and quotient rules, if they are easy to avoid.
(a) Rewrite f (x) using exponents to make its derivative easy to compute:
(b) Find f 0 (x)
(c) Find f 00 (x)
Solution: (a) First, distribute the x4 and then use properties of exponents
√
4
42 x
f (x) = x (πx) + x 15 − 1x4 = πx5 + 2x−21/2 − 1x4
x
(b) Now using basic rules for differentiation
f 0 (x) = 5πx4 − 21x−23/2 − 4x3
(c) The second derivative is
f 00 (x) = 20πx3 +
Example 8. Let f (x) =
√
3
12x −
x6
√
5
√ − x5 x. Find f 0 (x).
x
Solution: First, write
f (x) =
Then
483 −25/2
x
− 12x2
2
√
3
12x1/3 − 5x−13/2 − x11/2
√
3
12 −2/3 65 −15/2 11 9/2
x
+ x
− x
f (x) =
3
2
2
0
Example 9. This example illustrates the use of a graph to gain information about derivatives
of functions, or even their products and quotients.
Using the graph above, answer the following questions.
(a) At what values of x does f 0 (x) fail to exist?
(b) At what values of x does g 0 (x) fail to exist?
(c) Find: f (4), g(4), g 0 (4) and g 0 (4).
(d) Let h(x) = 6f (x) − 2g(x), find h0 (4).
(e) Let p(x) = f (x)g(x), find p0 (4).
(f) Let q(x) =
f (x)
, find q 0 (4).
g(x)
Solution: (a) f 0 (x) fails to exist when x = 3 and when x = 7 because the graph of f has
a cusp at those x values.
(b) g 0 (x) fails to exist when x = 3 and when x = 6 because g has a cusp (sharp corner) at
those values of x.
(c) From the graph f (4) = 7 and f 0 (4) = −1 [look at the slope of the graph of f at the
point (4, 7) to determine f 0 (4)].
From the graph g(4) = 3 and g 0 (4) = 0 [look at the slope of the graph of g at the point
(4, 7) to determine g 0 (4)].
(d) h0 (4) = 6f 0 (4) − 2g 0 (4) = (6)(−1) − (2)(0) = −6.0.
(f) Using the product rule: p0 (4) = f 0 (4)g(4) + g 0 (4)f (4) = (−1)(3) + (0)(7) = −3.0.
(g) Using the quotient rule
q 0 (4) =
(−1)(3) − (0)(7)
−3.00
f 0 (4)g(4) − g 0 (4)f (4)
=
=
[g(4)]2
(3)2
9.00
and the reader can simplify the fraction as appropriate.