EXAMPLE 4 Implicit Partial Differentiation
Find if the equation
defines z as a function of the two independent variables x and y and the partial derivative
exists.
Solution We differentiate both sides of the equation with respect to x, holding y constant
and treating z as a differentiable function of x:
EXAMPLE 5 Finding the Slope of a Surface in the y-Direction
The plane intersects the paraboloid in a parabola. Find the slope of the
tangent to the parabola at (1, 2, 5) (Figure 14.16).
Solution The slope is the value of the partial derivative at (1, 2):
0z
0y `
s1,2d
= 0
0y sx 2 + y 2d `
s1,2d
= 2y `
s1,2d
= 2s2d = 4.
0z>0y
x = 1 z = x 2 + y 2
0z
0x = z
yz - 1
.
A Function of Three Variables
If x, y, and z are independent variables and
then
EXAMPLE 7 Electrical Resistors in Parallel
If resistors of and ohms are connected in parallel to make an R-ohm resistor, the
value of R can be found from the equation
(Figure 14.17). Find the value of when and ohms.
Solution To find we treat and as constants and, using implicit differentiation,
differentiate both sides of the equation with respect to
When and
1
R = 1
30 + 1
45 + 1
90 = 3 + 2 + 1
90 = 6
90 = 1
15
,
R1 = 30, R2 = 45, R3 = 90,
0R
0R2
= R2
R2
2 = a R
R2
b
2
.
- 1
R2 0R
0R2
= 0 - 1
R2
2 + 0
0
0R2
a1
Rb = 0
0R2
a 1
R1
+ 1
R2
+ 1
R3
b
Partial Derivatives and Continuity
A function ƒ(x, y) can have partial derivatives with respect to both x and y at a point without
the function being continuous there. This is different from functions of a single variable,
where the existence of a derivative implies continuity. If the partial derivatives of
ƒ(x, y) exist and are continuous throughout a disk centered at however, then ƒ is
continuous at as we see at the end of this section.
EXAMPLE 8 Partials Exist, But ƒ Discontinuous
Let
(Figure 14.18).
(a) Find the limit of ƒ as (x, y) approaches (0, 0) along the line
(b) Prove that ƒ is not continuous at the origin.
(c) Show that both partial derivatives and exist at the origin.
Solution
(a) Since ƒ(x, y) is constantly zero along the line (except at the origin), we have
(b) Since the limit in part (a) proves that ƒ is not continuous at (0, 0).
(c) To find at (0, 0), we hold y fixed at Then for all x, and the
graph of ƒ is the line in Figure 14.18. The slope of this line at any xis In
particular, at (0, 0). Similarly, is the slope of line at any y, so
at (0, 0).
Example 8 notwithstanding, it is still true in higher dimensions that differentiability at
a point implies continuity. What Example 8 suggests is that we need a stronger requirement
for differentiability in higher dimensions than the mere existence of the partial derivatives.
We define differentiability for functions of two variables at the end of this section
and revisit the connection to continuity.
Second
and 02ƒ
0x0y
02ƒ
0y2 = 0
0y a0ƒ
0y b = -x cos y.
02ƒ
0x2 = 0
0x a0ƒ
0x b = ye x.
02ƒ
0x0y = 0
0x a0ƒ
0y b = -sin y + e x
02ƒ
0y0x = 0
0y a0ƒ
0x b = -sin y + e x
= cos y + ye x = -x sin y + e x
0ƒ
0y = 0
0y sx cos y + ye xd
0ƒ
0x = 0
0x sx cos y + ye xd
02ƒ
0x2 , 02ƒ
0y0x , 02ƒ
0y2 , and 02ƒ
0x0y .
ƒsx, yd = x cos y + yex,
ƒyx = sƒydx Means the same thing.
02ƒ
0x0y Differentiate first with respect to y, then with respect to x.
02ƒ
0x2 = 0
0x a0ƒ
0x b, 02ƒ
0x0y = 0
0x a0ƒ
0y b,
02ƒ
0y0x “d squared ƒdy dx” or ƒxy “ƒ sub xy”
02ƒ