The Chemistry Maths Book, Second Edition

4.6 Differentiation by rule 107

EXAMPLES 4.15The chain rule

(i) y 1 = 1 sin 12 x 1 = 1 sin 1 u, whereu 1 = 12 x

(ii)y 1 = 1 cos(2x

2

1 − 1 1) 1 = 1 cos 1 u, whereu 1 = 12 x

2

1 − 11

(iii) , whereu 1 = 12 x

2

1 − 11

(iv) y 1 = 1 ln(2x

2

1 − 1 1) 1 = 1 ln 1 u, whereu 1 = 12 x

2

1 − 11

(v)y 1 = 1 ln(sin 1 x) 1 = 1 ln 1 u, whereu 1 = 1 sin 1 x

0 Exercises 41–55

dy

dx

dy

du

du

dx u

x

x

x

=×=











 ==

1

(cos )

cos

cot

sin

xx

dy

dx

dy

du

du

dx u

x

x

x

=×=











×=

−

1

4

4

21

2

()

dy

dx

dy

du

du

dx

exxe

ux

=×= × =

−

()() 44

21

2

ye e

xu

==

21 −

2

dy

dx

dy

du

du

dx

=×=− × =−( sin ) ( )ux x x 4421 sin( −)

2

dy

dx

dy

du

du

dx

=×=(cos ) ( ) cosux×=22 2

Table 4.4 The chain rule

Type Function Derivative

power of uu

a

trigonometric sin 1 u

cos 1 u

tan 1 u

exponential e

u

logarithmic ln 1 u

1

u

du

dx

e

du

dx

u

sec

2

u

du

dx

−sinu

du

dx

cosu

du

dx

au

du

dx

a− 1