Calculus 12 – Ch. 2 Derivatives
Lesson 4: The Quotient Rule
THE QUOTIENT RULE
Given y =
f (x)
then
g(x)
y! =
or
if y =
g ( x ) f ! ( x ) " f ( x ) g! ( x )
#$ g ( x ) %&
2
f
gf ! " fg!
then y! =
g
g2
"Lo-dee-hi, hi-dee-lo, draw the line and square below"
(Lo being the denominator, Hi being the numerator and "dee" being the derivative.)
Examples
1) Find the derivative of each of the following using the quotient rule:
a) y =
y! =
=
=
x+2
x+3
b) y =
(x + 3)(1) " (x + 2)(1)
( x + 3)
x + 3" x " 2
( x + 3)
1
( x + 3)
2
2
2
y! =
=
=
x2 + 3
4!x
(4 " x)(2x) " (x 2 + 3)("1)
(4 " x)
2
8x " 2x 2 + x 2 + 3
(4 " x)
8x " x 2 + 3
(4 " x)
2
2
c) f (x) =
1
x
d) f (x) =
x+2
1!
x
1+ 2x
1 ! x !1
f (x) =
x+2
1
1 "2
x " x (2)
2
2
(1+ 2x )
(1+ 2x )
f !(x) =
1
x
=2
1
"
2
1
x
=2
1
"
2
=
1
2
+ x " 2x
(1+ 2x )
f "(x) =
1
2
2
=
(1+ 2x " 4x )
(1+ 2x )
=
2
1 " 2x
2 x (1+ 2x )
=
2
=
( x + 2) ( x !2 ) ! (1 ! x !1 ) (1)
2
( x + 2)
x !1 + 2x !2 !1+ x !1
( x + 2)
2x !1 + 2x !2 !1
( x + 2)
y! =
=
(x
3
+ x ) ( 0 ) " (1) ( 3x 2 +1)
(x
3
+ x)
2
"3x 2 "1
(x
3
f !(2) =
=
+ x)
2
"3(2)2 "1
((2)
3
+ (2))
"12 "1
(10)
2
13
="
100
2
1
10 = !13
x ! 2 100
!13x + 26 = 100y !10
y!
0 = 13x +100y ! 36
2
x !2 ( 2x + 2 ! x 2 )
( x + 2)
2
2x + 2 ! x 2
x 2 ( x + 2)
2
2) Find the equation of the tangent line to the curve y =
m = y!
2
! 1$
1
at
the
point
# 2, & .
" 10 %
x3 + x
f (x)
, where y(x) is the quotient of the functions u(x) and v(x) .
g(x)
3) Given y(x) =
Note:
Find y!(2) if:
f (2) = 3
f !(2) = 4
g(2) = 1
g!(2) = 2
!f$
f (x)
# & (x) =
g(x)
"g%
The derivative would be (using the quotient rule):
g(x) f !(x) " f (x)g!(x)
y!(x) =
[ g(x)]2
Then: y!(2) =
g(2) f !(2) " f (2)g!(2)
[ g(2)]2
(1) ( 4 ) # ( 3) (2)
2
[1]
! y"(2) =
= #2
x 2 !1
4) Find the points on the curve y = 4
where the tangents are horizontal.
x +1
Horizontal lines have slope of zero. Therefore, we need to find the derivative of the
given function (this will give us a function for the slope of any tangent line) and then solve
for the x-value(s) that make it equal to zero. We substitute the resulting x-value(s) into
the original function to find the corresponding y-value(s)
x =
2
m = y!
y! =
=
=
(x
4
)
(
)( )
+ 1 ( 2x ) " x " 1 4x
(x
2
)
+1
4
2
2x 5 + 2x " 4x 5 + 4x 3
(x
4
)
+1
2
3
0=
!2x 5 + 4x 3 + 2x
(x
0 = !2x ( x
)
+1
4
! 2x 2 ! 1
(x
4
)
+1
)
x 4 ! 2x 2 ! 1 = 0
"2x 5 + 4x 3 + 2x
2
!
2 $
Points of tangency: # ± 1 + 2 ,
& ( 0, !1)
4 + 2 2 &%
"
( !2 )2 ! 4(1)(!1)
2
2± 8
2
2±2 2
x2 =
2
2
x =1± 2
x2 =
2
4
" x = 0 or
!(!2) ±
"x = ± 1+ 2
!1
(
)
y=
"
(± 1 + 2 ) + 1
± 1+ 2
2
4
=
2
(1 + 2 )
2
+1
" x !1 %
5) Find the derivative of y = ( 2x 7 ! x 2 ) $
'
# x +1 &
# ( x +1) (1) " ( x "1) (1) &
# x "1 &
7
2
(
y! = (14x 6 " 2x ) %
( + ( 2x " x ) %%
2
(
$ x +1 '
( x +1)
$
'
(14x
=
6
" 2x ) ( x "1)
(2x
+
7
" x 2 ) ( x +1 " x +1)
2
( x +1)
( x +1)
(14x 6 " 2x ) ( x "1) ( x +1) + (2x 7 " x 2 ) (2)
=
2
( x +1)
(14x
=
=
6
" 2x ) ( x 2 "1) + 4x 7 " 2x 2
( x +1)
2
14x 8 "14x 6 " 2x 3 + 2x + 4x 7 " 2x 2
( x +1)
2
NOTE:
1)
2)
3)
4)
Slopes of zero (horizontal lines)
Undefined slope (vertical lines)
Parallel lines have the same (equal) slope
Perpendicular lines have negative reciprocal slopes
When simplifying derivatives:
1) Your final answer should only have positive exponents
2) Your final answer should not have any complex fractions
3) Factor and reduce if you can