PARTIAL DIFFERENTIATION 347G
[
It is noted that∂^2 z
∂x∂y=∂^2 z
∂y∂x]Problem 8. Show that when z=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−tsinθHence∂^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−tcosθHence∂^2 z
∂t∂θ=∂^2 z
∂θ∂tProblem 9. Show that if z=x
ylny, then(a)∂z
∂y=x∂^2 z
∂y∂xand (b) evaluate∂^2 z
∂y^2whenx=−3 andy=1.(a) To find
∂z
∂x,yis kept constant.Hence∂z
∂x=(
1
ylny)
d
dx(x)=1
ylnyTo 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)(2y)y^4⎫
⎪⎪
⎬⎪⎪
⎭using the quotient rule=x
y^4[−y− 2 y+ 2 ylny]=xy
y^4[− 3 +2lny]=x
y^3(2 lny−3)Whenx=−3 andy=1,∂^2 z
∂y^2=(−3)
(1)^3(2ln 1−3)=(−3)(−3)= 9Now try the following exercise.Exercise 140 Further problems on second
order partial derivativesIn Problems 1 to 4, find (a)∂^2 z
∂x^2(b)∂^2 z
∂y^2(c)∂^2 z
∂x∂y(d)∂^2 z
∂y∂x1.z=(2x− 3 y)^2[
(a) 8 (b) 18
(c)−12 (d) − 12]