Cambridge Additional Mathematics

Introduction to differential calculus (Chapter 13) 351

3 Find the gradient of the tangent to:

a y=

p

1 ¡x^2 at x=^12 b y=(3x+2)^6 at x=¡ 1

c y=

1

(2x¡1)^4

at x=1 d y=6£^3

p

1 ¡ 2 x at x=0

e y=^4

x+2

p

x

at x=4 f y=

³

x+^1

x

́ 3

at x=1.

4 The gradient function of f(x)=(2x¡b)a is f^0 (x)=24x^2 ¡ 24 x+6.

Find the constantsaandb.

5 Suppose y=

a

p

1+bx

whereaandbare constants. When x=3, y=1and

dy

dx

=¡^18.

Findaandb.

6 If y=x^3 then x=y

1

(^3).
a Find
dy
dx
and
dx
dy
, and hence show that
dy
dx
£
dx
dy
=1.
b Explain why
dy
dx
£
dx
dy
=1whenever these derivatives exist for any general function y=f(x).
We have seen the addition rule:
If f(x)=u(x)+v(x) then f^0 (x)=u^0 (x)+v^0 (x).
We now consider the case f(x)=u(x)v(x).Isf^0 (x)=u^0 (x)v^0 (x)?
In other words, does the derivative of a product of two functions equal the product of the derivatives of the
two functions?

Discovery 6 The product rule

Suppose u(x) and v(x) are two functions ofx, and that f(x)=u(x)v(x) is the product of these

functions.

The purpose of this Discovery is to find a rule for determining f^0 (x).

What to do:

1 Suppose u(x)=x and v(x)=x,sof(x)=x^2.

a Find f^0 (x) by direct differentiation. b Find u^0 (x) and v^0 (x).

c Does f^0 (x)=u^0 (x)v^0 (x)?

2 Suppose u(x)=x and v(x)=

p

x,sof(x)=x

p

x=x

3

(^2).
a Find f^0 (x) by direct differentiation. b Find u^0 (x) and v^0 (x).
c Does f^0 (x)=u^0 (x)v^0 (x)?

G The product rule

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_13\351CamAdd_13.cdr Tuesday, 7 January 2014 9:53:40 AM BRIAN