Chapter 35
Total differential, rates of
change and small changes
35.1 Total differential
In Chapter 34, partial differentiation is introduced for
the case where only one variable changes at a time,
the other variables being kept constant. In practice,
variables may all be changing at the same time.
If z = f (u, v, w, ...), then the total differential, dz,
is given by the sum of the separate partial differentials
of z,
i.e. dz =
∂z
∂u
du +
∂z
∂v
dv +
∂z
∂w
dw + · · ·
(1)
Problem 1. If z = f (x, y) and z = x 2 y 3 +
2x
y
+ 1,
determine the total differential, dz.
The total differential is the sum of the partial differentials,
i.e.
dz =
∂z
∂x
dx +
∂z
∂ y
dy
∂z
∂x
= 2x y
3
+
2
y
(i.e. y is kept constant)
∂z
∂ y
= 3x
2 y
2 2x
y 2 (i.e. x is kept constant)
Hence dz =
2xy
3
+
2
y
dx +
3x
2 y
2
−
2x
y 2
dy
Problem 2. If z = f (u, v, w) and
z =3u 2 − 2v + 4w 3 v 2 find the total differential, dz.
The total differential
dz =
∂z
∂u
du +
∂z
∂v
dv +
∂z
∂w
dw
∂z
∂u
= 6u (i.e. v and w are kept constant)
∂z
∂v
= −2 + 8w
3
v
(i.e. u and w are kept constant)
∂z
∂w
= 12w
2
v
2 (i.e. u and v are kept constant)
Hence
dz = 6u du + (8vw
3
− 2) dv + (12v
2 w
2 ) dw
Problem 3. The pressure p, volume V and
temperature T of a gas are related by pV = kT ,
where k is a constant. Determine the total
differentials (a) dp and (b) dT in terms of p, V
and T .
(a) Total differential dp =
∂ p
∂T
dT +
∂ p
∂V
dV .
Since pV = kT then p =
kT
V
hence
∂ p
∂T
=
k
V
and
∂ p
∂V
= −
kT
V 2
Thus dp =
k
V
dT −
kT
V 2 dV