The Chemistry Maths Book, Second Edition

110 Chapter 4Differentiation

EXAMPLE 4.18Inverse hyperbolic functions

If thenx 1 = 1 asinh 1 yand

Because cosh y 1 > 10 (see Figure 3.22), and

0 Exercises 63, 64

4.7 Implicit functions

Every functional relationship between two variables xand ycan be expressed in the

implicit form (see Section 2.4)

f(x, y) 1 = 10 (4.15)

In all the examples of the applications of the rules of differentiation discussed in

Section 4.6 it has been possible to express at least one of the variables as an explicit

function of the other. In some cases however neither variable can be so expressed and

the rules must be applied to the implicit function itself. Let equation (4.15) define

yas a function of x. Then the change in ythat accompanies a change in xis such

that equation (4.15) is always true. It follows that the rate of change of the implicit

functionf(x, y)is zero for all allowed changes:

(4.16)

EXAMPLE 4.19Finddy 2 dxforf(x, y) 1 = 1 y

5

1 − 12 y 1 − 1 x 1 = 10.

We have

=− −=()52 10

4

y

dy

dx

df

dx

d

dx

y

d

dx

y

d

dx

xy

dy

dx

dy

dx

=− −=−−() () ()

54

2521

d

dx

fxy(),= 0

dy

dx

ax

=

• 
  

1

22

aya yaxcosh =+ 1 sinh = +,

222

dx

dy

ay

dy

dx a y

=,=cosh

cosh

1

y

x

a

=













−

sinh

1