Partial

Derivatives

Chapter 2

Section 3

Let f be a function of two variables x and

y . The partial derivative of f with respect to x

is the function, denoted by

Definition.

such that its value at any point (x,y) in the

domain of f is given by

if this limit exists.

1D f 1f xff

x

10

, ,, lim

h

f x h y f x yD f x y

h

Let f be a function of two variables x and

y . The partial derivative of f with respect to y

is the function, denoted by

Definition.

such that its value at any point (x,y) in the

domain of f is given by

if this limit exists.

2D f 2f yff

y

20

, ,, lim

h

f x y h f x yD f x y

h

Given that .

Find and .

Exercise.

2, 3 5f x y xy y

Solution.

1 ,f x y 2 ,f x y

10

, ,, lim

h

f x h y f x yD f x y

h

2 2

0

3 5 3 5limh

x h y y xy y

h

0

3limh

hy

h

0lim 3h

y

3y

Given that .

Find and .

Exercise.

2, 3 5f x y xy y

Solution.

1 ,f x y 2 ,f x y

20

, ,, lim

h

f x y h f x yD f x y

h

2 2

0

3 5 3 5limh

x y h y h xy y

h

2

0

3 2limh

xh yh h

h

0lim 3 2h

x y h

3 2x y

REMARKS.

1.

2.

3.

4.

0 0 0 0

1 0 00

, ,, lim

h

f x h y f x yD f x y

h

0

0 0 01 0 0

0

, ,, lim

x x

f x y f x yD f x y

x x

0

0 0 02 0 0

0

, ,, lim

x x

f x y f x yD f x y

y y

0 0 0 0

2 0 00

, ,, lim

h

f x y h f x yD f x y

h

Let be a point in and f

be a function of n variables .

Definition.

The partial derivative of f with respect to xk is

the function, denoted by

such that its function value at any point P in

the domain of f is given by

if this limit exists.

1 2, ,..., nP x x xnR

1 2, ,..., nx x x

kD f

1 2

0

, ,..., ,...,lim

k nk

h

f x x x h x f PD f P

h

Find and

Let

Example.

Solution.

1 ,f x y 6xy

Differentiate with respect to x ;

Treating y as a constant. 4y

2 4, 3 7f x y x y xy

1 ,f x y 2 ,f x y

Find and

Let

Example.

Solution.

2 ,f x y 23x

Differentiate with respect to y ;

Treating x as a constant. 34xy

2 4, 3 7f x y x y xy

1 ,f x y 2 ,f x y

If is a differentiable function of and

defined by , where

and

all exist, then is a function of and

and

Chain Rule

u x y

,u f x y

, , ,x F r s y G r s , , ,x x y y

r s r s

u r s

,u u x u y

r x r y r

u u x u y

s x s y s

Find and .

Let

Exercise.

Solution.

2 33 4 , ,r

u x y x r ys

u

r

u

s

u

r

u

x

x

r

y

r

u

y

6x1

2 r

212y 1

s

Find and .

Let

Exercise.

Solution.

u

r

u

s

u

s

u

x

x

s

y

s

u

y

6x 0 212y 2

r

s

2 33 4 , ,r

u x y x r ys

Find and .

Let where

Exercise.

Solution.

2 2 2u x y z

,u u

u

u

u

x

x

y

u

y

sin cosx

sin siny cosz and

u

z

z

Find and

Let

Exercise.

Solution.

, 3sin tanx xf x y e xy Arc

y

1 ,f x y 2 ,f x y

1 ,f x y xe 3cos xy y

2

1

1x

y

1

y

Find and

Let

Exercise.

Solution.

, 3sin tanx xf x y e xy Arc

y

1 ,f x y 2 ,f x y

2 ,f x y 3cos xy x

2

1

1x

y

2

1

y

x

Find and

Let

Exercise.

Solution.

2 3 3 2 2, 2 lnyf x y x y x x y

1 ,f x y 2 ,f x y

1 ,f x y 2 3

1

2 x y 32x y

23 2 yx 2 2

1

x y

2x

Find and

Let

Exercise.

Solution.

2 3 3 2 2, 2 lnyf x y x y x x y

1 ,f x y 2 ,f x y

2 ,f x y 2 3

1

2 x y 2 23x y

3 2 ln 2yx 2 2

1

x y

2y

Find and

Let

Exercise.

Solution.

3, log sec tanf x y x y

1 ,f x y 2 ,f x y

1 ,f x y 1

sec tan ln 3x y sec tan tanx x y

2 ,f x y 2sec secx y1

sec tan ln3x y

Let S be a surface given by .

Geometric Interpretation.

Consider a plane given by and

which intersects S at a curve C.

Suppose

is on C.

xy

: slope of the

tangent line to the

curve C at P0.

,z f x y

0y y

0 0 0 0, ,P x y z

1 ,f x y

Let S be a surface given by .

Consider a plane given by and

which intersects S at a curve C.

Suppose

is on C.

xy

: slope of the

tangent line to the

curve C at P0.

Geometric Interpretation.

,z f x y

0x x

0 0 0 0, ,P x y z

2 ,f x y

Find .

Let where

Total Derivative

Solution.

ln xu x y ye

du

dz

csc , cotx z y z

du

dz

u

x

dx

dz

dy

dz

u

y

ln xy ye csc cotz z

xxe

y

2csc z

Implicit Differentiation

2 3tan 4 0xxy y x y

Let and y f x

Find . dy

dx

Theorem

If f is a differentiable function of the single

variable x such that y = f (x) and f is

defined implicitly by the equation

F (x, y) = 0, then if F is differentiable and

, then , 0yF x y

Implicit Differentiation

,

,

x

y

F x ydy

dx F x y

Higher

Order

Partial

Derivatives

When we differentiate a function

twice, we produce its second-order derivatives.

These derivatives are usually denoted by

,f x y

Second-Order Partial Derivative

2

2

f

x

xxf

2

2

f

y

yyf

2 f

x y

yxf

2 f

y x

xyf

2 f f f

x y x y

The defining equations are

2 f f f

y x y x

Second-Order Partial Derivative

Let 2 2 3 3 4w xy x y x y

Solution.

w

x

2y32xy 2 43x y

46xy32y2

2

w

x

26xy2y2w

y x

w

x x

w

y x

2 312x y

Second-Order Partial Derivative

Let 2 2 3 3 4w xy x y x y

Solution.

w

y

2xy 2 23x y 3 34x y

26x y2x2

2

w

y

26xy2y2w

x y

w

y y

w

x y

2 312x y

3 212x y

Second-Order Partial Derivative

Find all of its second order partial

derivatives.

Let , 3sin tanxf x y e xy Arc y

Solution.

,xf x y xe 3 cosy xy

,xxf x y 23 siny xyxe

,xyf x y 3cos xy 3 sinxy xy

Second-Order Partial Derivative

Let , 3sin tanxf x y e xy Arc y

Solution.

,yf x y 3 cosx xy

,yyf x y

2

1

1 y

23 sinx xy

22

12

1

y

y

,yxf x y 3 sinxy xy 3cos xy

Theorem. The Mixed Derivative Theorem

,f x y

Second-Order Partial Derivative

,a b

If and its partial derivatives

Are defined throughout an open region

containing a point and are all continuous at

, then

, , ,x y xy yxf f f f

,a b

, ,xy yxf a b f a b

Second-Order Partial Derivative

Verify that given 2 2f f

x y y x

, 2 ln lnxf x y x y y x

(¼ paper)

Remark.

3 f

x y x

Higher-Order Partial Derivative

There is no theoretical limit to how many times we

can differentiate a function as long as the

derivatives exist.

xyxf

3

2

f

x y

yxxf

4

2 2

f

y x

xxyyf

4

3

f

x y

yxxxf

2 2, cosf x y x y

Exercise.

Higher-Order Partial Derivative

Let . Find 3

2

f

x y

Solution.

yxxf

,yf x y 2 22 siny x y

,yxf x y 2 24 cosxy x y

,yxxf x y 2 24 cosy x y

2 2 28 sinx y x y