354 Introduction to differential calculus (Chapter 13)

Example 12 Self Tutor

Use the quotient rule to find

dy

dx

if:

a y=

1+3x

x^2 +1

b y=

p

x

(1¡ 2 x)^2

a y=

1+3x

x^2 +1

is a quotient with u=1+3x and v=x^2 +1

) u^0 =3 and v^0 =2x

Now

dy

dx

=

u^0 v¡uv^0

v^2

fquotient ruleg

=

3(x^2 +1)¡(1 + 3x)2x

(x^2 +1)^2

=

3 x^2 +3¡ 2 x¡ 6 x^2

(x^2 +1)^2

=^3 ¡^2 x¡^3 x

2

(x^2 +1)^2

b y=

p

x

(1¡ 2 x)^2

is a quotient with u=x

1

(^2) and v=(1¡ 2 x)^2
) u^0 =^12 x
¡^12
and v^0 = 2(1¡ 2 x)^1 £(¡2) fchain ruleg
=¡4(1¡ 2 x)
Now
dy
dx

u^0 v¡uv^0
v^2
fquotient ruleg

1
2 x
¡^12
(1¡ 2 x)^2 ¡x
1
(^2) £(¡4(1¡ 2 x))
(1¡ 2 x)^4

1
2 x
¡^12
(1¡ 2 x)^2 +4x
1
(^2) (1¡ 2 x)
(1¡ 2 x)^4

(1¡ 2 x)
h
1 ¡ 2 x
2 px+4
p
x
³
2
p
x
2 px
́i
(1¡ 2 x)^4
flook for common factorsg

1 ¡ 2 x+8x
2
p
x(1¡ 2 x)^3

6 x+1
2
p
x(1¡ 2 x)^3
3
Simplification of is often unnecessary,
especially if you simply want the gradient
of a tangent at a given point. In such cases,
substitute a value for without simplifying
the derivative function first.
^^
x
dy
dx
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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_13\354CamAdd_13.cdr Tuesday, 7 January 2014 9:54:04 AM BRIAN