Freshman Classes: Partial Derivative Review and Applied Problems
Examples are taken from Thomas’s Calculus textbook 10th Ed.
And lecture notes from Dr. Leisner’s Calc III class
Joe Trout and Brittany McNeill
Revised by Learning Center
Review: Partial Derivative Example
What is a partial derivative?
A partial derivative with respect to x at the point x0 , y 0  is the ordinary derivative of f x, y0 
with respect to x at the point x  x0 .
When are they used?
A partial derivative is used for functions with more than one independent variable.
How do you calculate it?
The partial derivative is taken exactly like a normal derivative, with one exception. All but one
independent variable are treated like a constant. If you take the derivative with respect to x,
then y is treated as a constant.
Example:
Let f x, y   x 2  3xy  y  1 . Find the values of
f
f
and
at the point (4, -5).
x
y
SOLUTION:
f
To find
, we treat y as a constant and differentiate with respect to x.
x
f
 2

x  3xy  y  1  2 x  3 y
x x

The value of
To find

f
at (4, -5) is 2(4) + 3(-5) = -7.
x
f
, we treat x as a constant and differentiate with respect to y:
y
f

 x 2  3xy  y  1  3x  1 .
y x
The value of
f
at (4, -5) is 3(4) + 1 = 13.
y
See chapter 11 section 3 of Thomas’s Calculus 10th Ed. for more examples.
Uses: There are many uses of a partial derivative.
First you can use it for gradient. What is a gradient?
A gradient is a vector obtained by evaluating the partial derivatives of a function at P0. Once it
is calculated from the partial derivatives, it can be used to calculate a directional derivative. This
directional derivative then tells rate of change of a function in a particular direction.
The gradient of f in Cartesian coordinates: f 
df ˆ df ˆ df ˆ
i
j
k
dx
dy
dz
Gradient example:
Find the derivative of f(x,y)=xey + cos(xy) at the point (2,0) in the direction of v=3i-4j.
First, calculate the gradient by taking the two partial derivatives.
fx=ey – y sin (xy),
fy= xey– x sin (xy)
Next, calculate the gradient at the point (2,0)
fx(2,0) = e0-0 = 1,
fy(2,0) = (2) e0 – (2)(0) = 2
So the gradient of f at (2, 0) is
f
( 2, 0 )
i2j
Next, calculate the direction of v by dividing by its length
u
v v 3 4
  i j
v 5 5 5
Finally, the derivative of f at (2,0) in the direction of v is:
(Du f )
(Du f )
( 2, 0)
 f
u
( 2, 0)
3 4
3 8
 (i  2 j )  ( i  j )    1
( 2, 0)
5 5
5 5
You can use partial derivatives to calculate error. To calculate error you need to know several
formulas:
True
Estimate
f   f  f 0 
df
Relative
change
f
f ( x0 , y 0 )
df
f ( x0 , y 0 )
Percent
change
f
 100%
f ( x0 , y 0 )
df
 100%
f ( x0 , y 0 )
Linearization
L  f ( x0 , y0 )  f x ( x0 , y0 )( x  x0 )  f y ( x0 , y0 )( y  y0 )
Total
differential
df  f x ( x0 , y0 )dx  f y ( x0 , y0 )dy  f z ( x0 , y0 )dz
Absolute
change
Error Example:
Find a reasonable square about the point r0 , h 0  = ( 5, 12) in which the value of V   r 2 h will
not vary by more than ± 0.1.
SOLUTION:
We approximate the variation ∆V by the differential dV :
2
dV  2 r0 h0 dr   r0 dh  2 (5)(12)dr   (5) 2 dh  120 dr  25 dh .
Since the region to which we are restricting to our attention is a square, we may set dh = dr to
get
dV  120 dr  25 dr  145 dr
So, how small must we take dr to be sure that | dV | is no larger than 0.1? To answer this
question, we start with the inequality
| dV | ≤ 0.1,
Express dV in terms of dr
| 145 dr | ≤ 0.1,
Solve for dr:
| dr | =
0.1
 2.1  10 4 . Round down to make sure dr isn’t too big.
145
Using dh  dr , the square we want is described by the inequalities
| r - 5 | ≤ 2.1x104 , | h – 12 | ≤ 2.1x104 .
As long as r and h stay in this square, we expect | dV | to be less than or equal to 0.1 and we
may expect ∆V to be the same size.
You can also use partial derivatives to estimate change in the resistance in wires.
Resistance Example:
The total resistance R of 2 resistors (x, y) connected in parallel is R 
xy
.
x y
Suppose x and y are measured to be 200 Ω (ohms) and 400 Ω, respectively. If x increases by 1
Ω and y decreases by 4 Ω, estimate the change in R using the total differential.
SOLUTION:
The total differential is dR  Rx ( x0 , y0 )dx  R y ( x0 , y0 )dy  Rz ( x0 , y0 )dz .
First find the partial derivative with respect to x.
Rx 
y ( x  y )  x( y )(1)
y2

.
( x  y) 2
( x  y) 2
Then to find the value of Rx, substitute 200 and 400 into x and y.
Rx 200,400 
400 2 4

600 2 9
Then find the partial derivative with respect to y.
R y (200,400) 
x2
x  y 2

200, 400
1
9
Then find the change in x and y (from the problem statement)
∆x = 1Ω, ∆y = - 4Ω. Then plug them into the formula for the differential:
R  Rx  x  R y  y 
4
1
 1    4  0 
9
9
So the change in the resistance is zero overall.
Practice Problems (solutions on separate sheet):
1. Find the gradient of the function f ( x, y)  ln( x 2  y 2 ) at the point (1,1)
2. Find the derivative of the function f ( x, y, z)  xy  yz  zx at the point (1,-1, 2) in the
direction A  3i  6 j  2k
1
3. The volume of a cone is given by the equation: V   r 2 h .
3
Use the total differential to estimate the change in volume if the height h increases from 10 to
10.1 cm and the radius r decreases from 12 to 11.95 cm.
 
4. Find the linearization of f ( x, y)  e 3 y sin(3x) at the point  ,0  .
6 
5. Three resistors are connected in parallel as shown:
xyz
. Find the percent change in R if x, y, and z increase by 1 Ω, 0
xy  xz  yz
Ω and -2 Ω, and x, y, z   1,2,3 Ω.
The resistance R 
Solutions to practice problems:
1. Partial derivatives: f x 
f y (1,1) 
2x
2y
, fy  2
2
x y
x  y2
2
2 1
2 1
 1 and f y (1,1)  2 2  1 , So f
2
1 1
1 1
2
(1,1)
i j
2. f x  y  z , f y  x  z and f z  y  x
So f x (1,  1, 2)  1  2  1 , f y (1,  1, 2)  1  2  3 and f z (1,  1, 2)  1  1  0
f
(1, 1, 2 )
 i  3 j , And u 
A A 3 6
2
  i j k
A 7 7 7
7
Thus
(Du f )
(Du f )
(1, 1, 2 )
 f
(1, 1, 2 )
u
3
6
2
3 18
 (i  3 j )  ( i  j  k )  
3
(1, 1, 2 )
7
7
7
7 7
2
3. Partial derivative with respect to r: Vr   r h , Partial derivative with respect to h:
3
1
Vh   r 2 . ∆h = .01, ∆r = - .05, Total differential = V  Vr r  Vh h  0.8
3
4. Partial derivative with respect to x: f x  3e e y cos(3x) ,
Partial derivative with respect to y: f y  3e e y sin(3x) .
 
 
 
Plug the point  ,0  into f x and f y . f x  , 0   0, f y  , 0   3 .
6 
6 
6 


L( x, y)  f x0 , y 0   f x ( x0 , y 0 )x  f y ( x0 , y0 )y  f x ( ,0)( x  x0 )  f y ( ,0)( y  y 0 )  3 y
6
6
5. Percent change =
df
f
 100 =
f ( x0 , y 0 )
f ( x0 , y 0 )
df  f x ( x0 , y0 )dx  f y ( x0 , y0 )dy  f z ( x0 , y0 )dz , R  f x x  f y y  f z z
fx 
yz xy  xz  yz   xyz  y  z 
62  3  6  65 66  30 36



2
2
121
121
xy  xz  yz 


2

3

6
1, 2,3 
fy 
xz xy  xz  yz   xyz x  z 
32  3  6  64 33  24
9



2
2
121
121
xy  xz  yz 
2  3  6
1, 2,3 
fz 
xy xy  xz  yz   xyz x  y 
22  3  6  63 22  18
4



2
2
121
121
xy  xz  yz 
2  3  6
1, 2,3 
R  f x x  f y y  f z z 
36
4
28
2

 0.23  23 %
121
121 121