LECTURE 5: PARTIAL DERIVATIVES
MINGFENG ZHAO
January 14, 2015
Example 1. Let f (x, y) = cos(exy ), find fx and fy .
In fact, we have
fx (x, y)
= − sin(exy ) · exy · y
By the Chain rule
= −yexy sin(exy )
fy (x, y)
= − sin(exy ) · exy · x
By the Chain rule
= −xexy sin(exy ).
Recall the quotient rule for derivative of function of one variable:
0
f
f 0 g − g0 f
=
.
g
g2
Example 2. Let f (x, y) =
sin(x2 y)
, find fx and fy .
x − 2y
In fact, we have
fx (x, y)
fy (x, y)
=
cos(x2 y) · 2xy · (x − 2y) − sin(x2 y)
(x − 2y)2
=
2xy(x − 2y) cos(x2 y) − sin(x2 y)
(x − 2y)2
=
cos(x2 y) · x2 · (x − 2y) − sin(x2 y) · (−2)
(x − 2y)2
=
x2 (x − 2y) cos(x2 y) + 2 sin(x2 y)
.
(x − 2y)2
1
By the quotient rule and Chain rule
By the quotient rule and Chain rule
2
MINGFENG ZHAO
Higher order partial derivatives
Notation 1
∂ ∂f
∂2f
=
∂x ∂x
∂x2
∂ ∂f
∂2f
=
∂y ∂x
∂y∂x
∂2f
∂ ∂f
=
∂x ∂y
∂x∂y
∂2f
∂ ∂f
=
∂y ∂y
∂y 2
Notation 2
What we say...
(fx )x = fxx
f −x−x
(fx )y = fxy
f −x−y
(fy )x = fyx
f −y−x
(fy )y = fyy
f −y−y
Example 3. Let f (x, y) = 3x4 y − 2xy + 5xy 3 , find the four second order partial derivatives of f .
For the first order partial derivatives, we have
fx = 12x3 y − 2y + 5y 3 ,
fy = 3x4 − 2x + 15xy 2 .
Then
fxx
=
(fx )x = 36x2 y
fxy
=
(fx )y = 12x3 − 2 + 15y 2
fyx
=
(fy )x = 12x3 − 2 + 15y 2
fyy
=
(fy )y = 30xy.
Theorem 1 (Clairaut). fxy = fyx .
Remark 1. Clairuat’s theorem just says that we can switch the orders when computing partial derivatives.
Example 4. Let f (x, y) = 2y and g(x, y) = 3x, can you find a function F (x, y) such that Fx = f and Fy = g?
If such F exists, by Clairaut’s theorem, we have Fxy = Fyx . Since Fx = f , then Fxy = fy = 2. Since Fy = g, then
Fyx = gx = 3. So Fxy = 2 6= 3 = Fyx , contradiction. Therefore, there is no F (x, y) such that Fx = f and Fy = g.
Remark 2. If fy (x, y) = gx (x, y), then there exists a h(x, y) such that hx = f and hy = g. For example, let f (x, y) = 2y
and g(x, y) = 2x, then fy = 2 = gx . Let h(x, y) = 2xy, then hx (x, y) = 2y and hy (x, y) = 2x.
Example 5. Let f (x, y, z) = e−xy cos(z), find fx , fz , fxz and fxyz .
LECTURE 5: PARTIAL DERIVATIVES
3
In fact, we have
fx
= e−xy · (−y) · cos(z)
By the Chain rule
= −y cos(z)e−xy
fz
fxz
= −e−xy sin(z)
=
(fx )z
= y sin(z)e−xy
fxyz
=
(fxz )y
=
sin(z)e−xy + y sin(z)e−xy · (−x)
=
sin(z)e−xy − xy sin(z)e−xy
By the product rule and the Chain rule
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca