Cambridge Additional Mathematics

Introduction to differential calculus (Chapter 13) 353

2 Find

dy

dx

using the product rule:

a y=x^2 (2x¡1) b y=4x(2x+1)^3 c y=x^2

p

3 ¡x

d y=

p

x(x¡3)^2 e y=5x^2 (3x^2 ¡1)^2 f y=

p

x(x¡x^2 )^3

3 Find the gradient of the tangent to:

a y=x^4 (1¡ 2 x)^2 at x=¡ 1 b y=

p

x(x^2 ¡x+1)^2 at x=4

c y=x

p

1 ¡ 2 x at x=¡ 4 d y=x^3

p

5 ¡x^2 at x=1.

4 Consider y=

p

x(3¡x)^2.

a Show that

dy

dx

=

(3¡x)(3¡ 5 x)

2

p

x

b Find thex-coordinates of all points on y=

p

x(3¡x)^2 where the tangent is horizontal.

c For what values ofxis

dy

dx

undefined?

5 Suppose y=¡ 2 x^2 (x+4). For what values ofxdoes

dy

dx

=10?

Expressions like

x^2 +1

2 x¡ 5

,

p

x

1 ¡ 3 x

, and

x^3

(x¡x^2 )^4

are calledquotientsbecause they represent the division

of one function by another.

Quotient functions have the form Q(x)=

u(x)

v(x)

Notice that u(x)=Q(x)v(x)

) u^0 (x)=Q^0 (x)v(x)+Q(x)v^0 (x) fproduct ruleg

) u^0 (x)¡Q(x)v^0 (x)=Q^0 (x)v(x)

) Q^0 (x)v(x)=u^0 (x)¡

u(x)

v(x)

v^0 (x)

) Q^0 (x)v(x)=

u^0 (x)v(x)¡u(x)v^0 (x)

v(x)

) Q^0 (x)=

u^0 (x)v(x)¡u(x)v^0 (x)

[v(x)]^2

when this exists.

THE QUOTIENT RULE

H The quotient rule

If Q(x)=

u(x)

v(x)

then Q^0 (x)=

u^0 (x)v(x)¡u(x)v^0 (x)

[v(x)]^2

Alternatively, if y=

u

v

whereuandvare functions ofx, then

dy

dx

=

u^0 v¡uv^0

v^2

.

.

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_13\353CamAdd_13.cdr Monday, 20 January 2014 3:51:15 PM BRIAN