The Chemistry Maths Book, Second Edition

4.6 Differentiation by rule 103

EXAMPLES 4.9Differentiating powers

(i) y 1 = 1 x

5

(ii)f(x) 1 = 1 x

− 122

(iii)f(x) 1 = 1 x

0.3

0 Exercises 23–26

EXAMPLE 4.10For the ideal gas example discussed in Section 4.1,

sincenRTis constant at constant Tand n. We note that whereas Vis inversely

proportional to p, it is directly proportional to 12 p; that is,Vis a linear function of

12 p, and the graph ofVagainst 12 pis a straight line with slope

.

The rules for differentiating combinations of elementary functions are summarized

in Table 4.3; in these rules, xis the independent variable, y, uand vare functions of x,

and ais a constant.

dV

dp

nRT

()1

=

V

nRT

p

dV

dp

nRT

d

dp p

nRT

p

=, =













=−

















=

11

2

−−nRT

p

2

d

dx

fx x()=.

−.

03

07

′ =−

−

fx() x

1

2

32

dy

dx

= 5 x

4

Table 4.3 Differentiation of combinations of functions

Type Rule

1. multiple of a function


2. sum of functions


3. product rule


4. quotient rule


5. chain rule


6. inverse rule or

dx

dy

dy

dx

×= 1

dx

dy

dy

dx

=











1 

d

dx

fu

df

du

du

dx

()=×

d

dx

udu

dx

u

d

v dx

v

v

v











=−













2

d

dx

uu

d

dx

du

dx

()v

v

=+v

d

dx

u

du

dx

d

dx

()+= +v

v

d

dx

au a

du

dx

()=