Multivariate Calculus - Bruce E. Shapiro

64 LECTURE 9. THE PARTIAL DERIVATIVE
Similarly, by the chain rule
∂u
∂x = ∂ ( 1√t
e
/(4ct))
−x2
∂x
= √ 1 ( )
e −x2 /(4ct) −2x
t 4ct
By the product rule,
∂ 2 u
∂x 2 = − ∂
∂x
−x
=
2ct √ t
−x
=
2ct √ t
= 1 √
t
e −x2 /(4ct)
= √ 1 e −x2 /(4ct) ∂ ( ) −x
2
t ∂x 4ct
= − x
2ct √ t e−x2 /(4ct)
( )
x
2ct √ /(4ct)
t e−x2
∂ /(4ct)
∂x e−x2 − e −x2 /(4ct) ∂ x
∂x 2ct √ t
( ) ( )
e −x2 /(4ct) −2x
+
4ct
( x
2
4c 2 t 2 − 1
2ct
Multiplying the last equation through by c,
c ∂2 u
∂x 2 = √ 1 (
e −x2 /(4ct) x
2
t 4ct 2 − 1 )
2t
Equivalence of Mixed Partials
)
(
−e −x2 /(4ct) ) ( 1
2ct √ t
= ∂u
∂t
We observed at the end of example (9.5) that f xy = f yx . This property is true in
general, although it is not stated formally in the book until section 15.3 theorem B.
To see why it is true we observe the following:
f xy (x, y) = ∂ ∂y
∂f(x, y)
∂x
= ∂ ∂y f x(x, y)
= lim
k→0
f x (x, y + k) − f x (x, y)
k
f(x + h, y + k) − f(x, y + k)
lim
= lim h→0 h
k→0 k
)
− lim
h→0
f(x + h, y) − f(x, y)
h
= lim
k→0
lim
h→0
[f(x + h, y + k) − f(x, y + k)] − [f(x + h, y) − f(x, y)]
hk
[f(x + h, y + k) − f(x + h, y)] − [f(x, y + k) − f(x, y)]
= lim lim
k→0 h→0 hk
f(x + h, y + k) − f(x + h, y)
lim
= lim k→0 k
h→0
h
f y (x + h, y) − f y (x, y)
= lim
h→0 h
= ∂
∂x f y(x, y) = f yx (x, y).
Hence in general it is safe to assume that f xy = f yx .
− lim
k→0
f(x, y + k) − f(x, y)
k
Revised December 6, 2006. Math 250, Fall 2006