112 Chapter 4Differentiation
and, multiplying by y,
(ii)y 1 = 1 a
x
,ln 1 y 1 = 1 x 1 ln 1 a.
Then
(iii)y 1 = 1 x
x
,ln 1 y 1 = 1 x 1 ln 1 x.
Applying the product rule,
and, therefore,
0 Exercises 69–72
EXAMPLE 4.21Logarithmic plots
For a system undergoing exponential decay (first-order kinetics), the size of the
system is given by (see Example 3.20)
x(t) 1 = 1 x
0
e
−kt
and, taking the logarithm of both sides,
ln 1 x 1 = 1 ln 1 x
0
1 − 1 kt
A plot of ln xagainst t, Figure 4.8, gives a straight line with
sloped(ln 1 x) 2 dt 1 = 1 −k, and intercept ln 1 x
0
with the axist 1 = 10.
This example is important because it demonstrates the
standard way of calculating the rate constant kfrom a linear
plot of experimental values of xand t.
0 Exercise 73
1
11
y
dy
dx
x
dy
dx
xx
x
=+ln and =( ln )+
d
dx
yx
d
dx
xx
d
dx
ln =+=+ln ln xx 1 ln
d
dx
y
y
dy
dx
a
dy
dx
aa
x
ln ==ln , ,=ln
1
and therefore
dy
dx
x
x
x
xx
=
−
−
=
+−
1
1
1
1
1
11
2
12
12 32
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Figure 4.8