17. L ECTURE 17
Objectives
⇤ Review Partial Derivatives
For Lecture 17, the following exercises were done in class.
17.1. Partial Derivatives.
(1) Find fx and fy for
f (x, y) = xy 2
x3 y.
To find fx , we treat x as a variable and y as a constant. Therefore,
fx (x, y) = y 2
3x2 y
By the same logic,
fy (x, y) = 2xy
(2) Find fx and fy for
f (x, y) = ln(x +
x3
p
x2 + y 2 ).
To find fx , we treat x as a variable and y as a constant. Here, we will need to use the
chain rule twice. The “outside” function is the natural log function. The middle function is
x plus the square root function. Therefore,
1
1
p
fx (x, y) =
· (1 + (x2 + y 2 )
2
x + x2 + y 2
1
1
2
1 + x(x2 + y 2 ) 2
p
· 2x) =
x + x2 + y 2
The approach for fy is the same, except x is treated as a constant.
1
1
p
fy (x, y) =
· (0 + (x2 + y 2 )
2
x + x2 + y 2
97
1
1
2
y(x2 + y 2 ) 2
p
· 2y) =
x + x2 + y 2
(3) Find fx , fy , and fz for
5x2 y 3 z 4 .
f (x, y, z) = xz
fx (x, y) = z
10xy 3 z 4
fy (x, y) = 0
15x2 y 2 z 4
fz (x, y) = x
20x2 y 3 z 3
(4) Find fxxx and fxyx for
f (x, y) = x4 y 2
fx (x, y) = 4x3 y 2
x3 y.
3x2 y
fxx (x, y) = 12x2 y 2
6xy
fxxx (x, y) = 24xy 2
6y
fxyx (x, y) = fxxy (x, y) = 24x2 y
98
6x
(5) Find zuvw for
p
z=u v
zuvw =
w.
p
zu = v w
1
1
zuv = (v w) 2 · 1
2
3
1
1
(v w) 2 · ( 1) = (v
4
4
(6) Find wzyx and wxxy for
w=
w)
3
2
x
.
y + 2z
1
y + 2z
= ( 1)(y + 2z)
wx =
wxy
wzyx = wxyz = ( 1)( 2)(y + 2z)
3
wxxy = wxyx = 0
99
2
·1
· 2 = 4(y + 2z)
3
P ENN S TATE U NIVERSITY, U NIVERSITY PARK , PA 16802
85