Cambridge Additional Mathematics

352 Introduction to differential calculus (Chapter 13)

3 Copy and complete the following table, finding f^0 (x) by direct differentiation.

f(x) f^0 (x) u(x) v(x) u^0 (x) v^0 (x) u^0 (x)v(x)+u(x)v^0 (x)

x^2 x x

x

3

(^2) x px
x(x+1) x x+1
(x¡1)(2¡x^2 ) x¡ 1 2 ¡x^2
4 Copy and complete: “If f(x)=u(x)v(x) then f^0 (x)=::::::”

THE PRODUCT RULE

If f(x)=u(x)v(x) then f^0 (x)=u^0 (x)v(x)+u(x)v^0 (x).

Alternatively, if y=uv whereuandvare functions ofx, then

dy

dx

=

du

dx

v+u

dv

dx

.

Example 11 Self Tutor

Find

dy

dx

if:

a y=

p

x(2x+1)^3 b y=x^2 (x^2 ¡ 2 x)^4

a y=

p

x(2x+1)^3 is the product of u=x

1

(^2) and v=(2x+1)^3
) u^0 =^12 x
¡^12
and v^0 = 3(2x+1)^2 £ 2 fchain ruleg
= 6(2x+1)^2
Now
dy
dx
=u^0 v+uv^0 fproduct ruleg
=^12 x
¡^12
(2x+1)^3 +x
1
(^2) £6(2x+1)^2
=^12 x
¡^12
(2x+1)^3 +6x
1
(^2) (2x+1)^2
b y=x^2 (x^2 ¡ 2 x)^4 is the product of u=x^2 and v=(x^2 ¡ 2 x)^4
) u^0 =2x and v^0 =4(x^2 ¡ 2 x)^3 (2x¡2) fchain ruleg
Now
dy
dx
=u^0 v+uv^0 fproduct ruleg
=2x(x^2 ¡ 2 x)^4 +x^2 £4(x^2 ¡ 2 x)^3 (2x¡2)
=2x(x^2 ¡ 2 x)^4 +4x^2 (x^2 ¡ 2 x)^3 (2x¡2)

EXERCISE 13G

1 Use the product rule to differentiate:

a f(x)=x(x¡1) b f(x)=2x(x+1) c f(x)=x^2

p

x+1

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_13\352CamAdd_13.cdr Tuesday, 7 January 2014 9:53:47 AM BRIAN