Product and
Quotient
Rules
4.3
A particle moves along a line so that at time t where t  0
it’s position is given by s ( t )  2 e  t . What is the velocity
t
2
of the particle when its acceleration is zero?

Find the derivative using the power rule. Simplify first.
y x x
2
5

product rule:
d
dx
 uv   u
dv
dx
v
du
dx
Notice that this is not just the
product of two derivatives.
This is sometimes memorized as: d  u v   u d v  v d u

Find the derivative.

y x 4
2
x
d uv   u dv  v du
4
 2x 1


quotient rule:
y
x
5
x
2
 u  v du  u dv
d 
2
v
v

quotient rule:
 u  v du  u dv
d 
2
v
v
 
x 3
2
y
x  5x  1
4

quotient rule:
 u  v du  u dv
d 
2
v
v
 
d 2x  5x
3
dx
x 3
2

Is the quotient rule necessary for this problem?
x  2x  4x
3
y 
2
x

Use your calculator to graph the function and it’s derivative.
y  sin x
d
sin x  cos x
dx

Use your calculator to graph the function and it’s derivative.
y  cos x
d
cos x   sin x
dx

We can find the derivative of tangent x by using the
quotient rule.
d
cos x  sin x
2
tan x
2
2
cos x
dx
d sin x
1
dx cos x
cos x
2
cos x  cos x  sin x    sin x 
2
sec x
2
cos x
d
dx
tan  x   sec x
2

Derivatives of the remaining trig functions can be
determined the same way.
d
d
sin x  cos x
cos x   sin x
dx
d
dx
2
dx
dx
d
cot x   csc x
d
sec x  sec x  tan x
dx
tan x  sec x
2
d
csc x   csc x  cot x
dx
p
P215 #1-31odd,
33a,c,d,35,37,39