Partial differentiation 349
(c)
∂^2 z
∂x∂y
=
∂
∂x
(
∂z
∂y
)
=
∂
∂x
( 12 x^2 y^2 + 14 y)= 24 xy^2
(d)
∂^2 z
∂y∂x
=
∂
∂y
(
∂z
∂x
)
=
∂
∂y
( 8 xy^3 − 6 x^2 )= 24 xy^2
[
It is noted that
∂^2 z
∂x∂y
=
∂^2 z
∂y∂x
]
Problem 8. Show that whenz=e−tsinθ,
(a)
∂^2 z
∂t^2
=−
∂^2 z
∂θ^2
,and(b)
∂^2 z
∂t∂θ
=
∂^2 z
∂θ∂t
(a)
∂z
∂t
=−e−tsinθand
∂^2 z
∂t^2
=e−tsinθ
∂z
∂θ
=e−tcosθand
∂^2 z
∂θ^2
=−e−tsinθ
Hence
∂^2 z
∂t^2
=−
∂^2 z
∂θ^2
(b)
∂^2 z
∂t∂θ
=
∂
∂t
(
∂z
∂θ
)
=
∂
∂t
(e−tcosθ)
=−e−tcosθ
∂^2 z
∂θ∂t
=
∂
∂θ
(
∂z
∂t
)
=
∂
∂θ
(−e−tsinθ)
=−e−tcosθ
Hence
∂^2 z
∂t∂θ
=
∂^2 z
∂θ∂t
Problem 9. Show that ifz=
x
y
lny,then
(a)
∂z
∂y
=x
∂^2 z
∂y∂x
and (b) evaluate
∂^2 z
∂y^2
when
x=−3andy=1.
(a) To find
∂z
∂x
,yis kept constant.
Hence
∂z
∂x
=
(
1
y
lny
)
d
dx
(x)=
1
y
lny
To find
∂z
∂y
,xis kept constant.
Hence
∂z
∂y
=(x)
d
dy
(
lny
y
)
=(x)
⎧
⎪⎪
⎨
⎪⎪
⎩
(y)
(
1
y
)
−(lny)( 1 )
y^2
⎫
⎪⎪
⎬
⎪⎪
⎭
using the quotient rule
=x
(
1 −lny
y^2
)
=
x
y^2
( 1 −lny)
∂^2 z
∂y∂x
=
∂
∂y
(
∂z
∂x
)
=
∂
∂y
(
lny
y
)
=
(y)
(
1
y
)
−(lny)( 1 )
y^2
using the quotient rule
=
1
y^2
( 1 −lny)
Hencex
∂^2 z
∂y∂x
=
x
y^2
(1−lny)=
∂z
∂y
(b)
∂^2 z
∂y^2
=
∂
∂y
(
∂z
∂y
)
=
∂
∂y
{
x
y^2
( 1 −lny)
}
=(x)
d
dy
(
1 −lny
y^2
)
=(x)
⎧
⎪⎪
⎨
⎪⎪
⎩
(y^2 )
(
−
1
y
)
−( 1 −lny)( 2 y)
y^4
⎫
⎪⎪
⎬
⎪⎪
⎭
using the quotient rule
=
x
y^4
[−y− 2 y+ 2 ylny]
=
xy
y^4
[− 3 +2lny]=
x
y^3
(2lny− 3 )
Whenx=−3andy=1,
∂^2 z
∂y^2
=
(− 3 )
( 1 )^3
(2ln1− 3 )=(− 3 )(− 3 )= 9
Now try the following exercise
Exercise 139 Further problemson second
order partial derivatives
In Problems 1 to 4, find (a)
∂^2 z
∂x^2
(b)
∂^2 z
∂y^2
(c)
∂^2 z
∂x∂y
(d)
∂^2 z
∂y∂x
- z=( 2 x− 3 y)^2
[
(a) 8 (b) 18
(c)− 12 (d)− 12
]