25 Partial Derivatives

8/13/2019 25 Partial Derivatives

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Partial Derivatives Exercises

Partial DerivativesMathematics 54 - Elementary Analysis 2

Institute of MathematicsUniversity of the Philippines-Diliman

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Partial DerivativesMathematics 54 - Elementary Analysis 2

Institute of MathematicsUniversity of the Philippines-Diliman

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Partial Derivatives Exercises Higher Order Partial Derivatives

Introduction

Let z

= f (x , y ) be dened at ( a , b ). If we x y

=b , then f (x , b ) becomes a

function of x alone.

The slope of the above tangent line to this curve at ( a , b ) is given by d

dx f (x , b )

x =a =: f x (a , b ).

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Introduction

Similarly, if we x x =a , then f (a , y ) becomes a function of y alone.

The slope of the above tangent line to this curve at ( a , b ) is given by d dy

f (a , y ) y =b =

: f y (a , b ).

These derivatives will be how differentiation on multivariate functions is

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Partial Derivatives

Denition

Let z = f (x , y ).1 The partial derivative of f with respect to x at any point ( x , y ) is

f x (x , y ) = lim x 0 f (x

+ x , y )

f (x , y )

x ,

provided that this limit exists.2 The partial derivative of f with respect to y at any point ( x , y ) is

f y (x , y ) = lim y 0 f (x , y

+ y )

f (x , y )

y ,

provided that this limit exists.

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l h d l


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Partial Derivatives

Example

Find f x (x , y ) given that f (x , y ) =2x 2 y

2

+2 y +4x .Solution. Consider

f (x + x , y ) f (x , y ) x =

2(x + x )2 y 2 +2( y )+4(x + x ) (2x

2 y 2 +2 y +4x ) x

=2x 2 y 2

+4x ( x ) y 2

+2( x )2 y 2

+2 y

+4x

+4 x

2x 2 y 2

2 y

4x

x

=4x ( x ) y 2 +2( x )

2 y 2 +4 x x

= 4xy 2+2 y

2 x +4

Thus,

f x (x , y ) = lim x 0(4xy 2 +2 y

2 x +4)

= 4xy 2+4.

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Partial Derivatives

Remark.

The partial derivatives of f at (a , b ) are also given by

1 f x (a , b ) = limx a f (x , b ) f (a , b )

x a 2 f y (a , b ) = lim y b

f (a , y ) f (a , b ) y b

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Partial Derivatives

Example

Find f x (1,1) and f y (1,1) given that f (x , y ) = 3x y +4xy 2 .

Solution. For f x , let us use the denition. We have

f (1 + x ,1) f (1,1) x =

3(1

+ x )

1 +4(1 + x )(1)2

3(1)

1 +4(1)(1)2

x

=3 +3 x +4 x 3

x

=7 x x

= 7.

Thus, f x (1,1)

= lim x

0

7

=7.

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Example. (continued...)

Find f x (1,1) and f y (1,1) given that f (x , y ) =3x y +4xy

2 .

For f y , let us use the previous remark. Consider

f (1, y ) f (1,1) y

1 =

3 y +4 y

27

y

1

=

3 +4 y 37 y

y y 1

=

(2 y +3)(2 y 1)( y 1) y ( y 1)

=(2 y +3)(2 y 1)

y .

Hence,

f y (1,1) =lim y 1(2 y

+3)(2 y

1)

y =5. 9/20



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Partial Derivatives

Remark.The partial derivative with respect to a variable can also be obtained by applying theorems for ordinary differentiation while treating the othervariables constant.

ExampleDetermine the partial derivative of f (x , y ) =2x

2 y 2 +2 y +4x with respect tox .

f x (x , y ) = (2 y 2) d

dx (x 2)+2 y d dx (1) +4 d dx (x )

= 4xy 2+4.

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g

Partial Derivatives

Example

Find f x (1,1) and f y (1,1) given that f (x , y ) =3x y +4xy

2 .

First, f x (x , y ) =3 y

d dx

(x )+4 y 2 d

dx (x ) =

3 y +4 y

2 .

Thus, f x (1,1) =31 +4(1)

2=7.

Similarly, f y (x , y ) =3x d dy

1 y +4x

d dy ( y 2) =

3x y 2 +8xy .

Hence, f y (1,1) =3(1)12 +8(1)(1) =5.

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Partial Derivatives

Notations.Let z = f (x , y ). The following denotes the partial derivatives of f withrespect to x :

f x (x , y )

D x f (x , y )

x f (x , y )

z x

Similarly, the partial derivatives of f with respect to y are denoted by:

f y (x , y )

D y f (x , y )

y f (x , y )

z y

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Partial Derivatives

Example

Evaluate

y sin 2(2x 3 y ).

Solution .

y

sin 2(2x 3 y ) = y sin(2 x 3 y ) 2

= 2sin(2 x 3 y )cos(2 x 3 y )(3)= 6sin(2 x 3 y )cos(2 x 3 y )

Note.

f x and f y are also called the derivatives of f in the x - and y -directions,respectively. Moreoever, they can also be interpreted as rates of change.

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Partial Derivatives in Functions of n Variables

Remark.

For a function of n variables x 1 , x 2 , . . . , x n , there are n partial derivatives.The partial derivative with respect to the i th variable is

f x i (x 1 , x 2 , . . . , x n ) = lim x i 0 f (x 1 , . . . , x i 1 , x i + x i , x i +1 , . . . , x n ) f (x 1 , . . . , x n )

x i .

Also, f x i (x 1 , x 2 , . . . , x n ) is obtained by performing ordinary differentiation with respect to x i while treating the other variables constant.

Notation.

Let u = f (x 1 , x 2 , . . . , x n ). The following are the other notations for the partialderivative with respect to x i :D x i f (x 1 , x 2 , . . . , x n ) u x i

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Partial Derivatives of Functions of n Variables

Example

Let f (x , y , z ) =xz y 2 +sin(2 x +3 y +4z )+e

x 2z . Find f x , f y and f z .

Solution. f x (x , y , z ) =

z y 2 +2cos(2 x +3 y +4z )+2xze

x 2z

f y (x , y , z ) = 2xz

y 3 +3cos(2 x +3 y +4z )

f z (x , y , z ) = x y 2 +4cos(2 x +3 y +4z )+x 2e x 2z

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Higher Order Partial Derivatives

Let z = f (x , y ). The functions f x (x , y ) and f y (x , y ) are also functions of x and y .

These functions also have partial derivatives. In fact, there are four of them.

x

z

x =

2z

x 2

= f xx

y

z x =

2z y x = f xy

x

z y =

2z x y = f yx

y z y =

2z y 2 = f yy

The above derivatives are called the second order partial derivatives of f .Remark. Analogous denitions can be made for functions of three or

more variables and for third or higher order partial derivatives.16/20



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Example

Let w = f (x , y ) =ln(2 x +3 y ). Find (a) 2w x y

and (b) f xy (x , y ).

Solution . (a) w

y = f y (x , y ) =1

2x

+3 y

(3) =3

2x

+3 y

2w x y = f yx (x , y ) =

x 3(2x +3 y )

1=(3)(2x +3 y )

2(2) = 6

(2x +3 y )2

(b) w

x =

f x (x , y )

=

1

2x +3 y (2)

=

2

2x +3 y 2w y x = f xy (x , y ) =

y 2(2x +3 y )

1=(2)(2x +3 y )

2(3) = 6

(2x +3 y )2Remark. The above example shows that f xy = f yx .

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Clairauts Theorem

Let f be a function of x and y . If f xy and f yx are continuous on somecircular region R on a plane, then for all ( x , y ) on R ,

f xy = f yx .

Example

No function f (x , y ) exists such that f x (x , y ) =3x +2 y and f y (x , y ) =4x +5 y .

Indeed, if such a function exists, then f xy

= f yx . But f xy (x , y )

=2 and

f yx (x , y ) =4, which are both continuous, yet are not equal.

Remark.

Clairauts theorem can also be extended to functions of n variables, n

N .

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Example

Let f (x , y , z ) =z ln( x 2 y cos z ). Find 3 f

x x z and f xzx .

Solution.

f (x , y , z )

z = ln(x 2 y cos z )

+z

1

x 2 y cos z x 2 y (

sin z )

= 2 ln x +ln y +ln(cos z )z tan z

2 f (x , y , z ) x z =

2x

3 f (x , y , z ) x x z =

2x 2

For f xzx , recall that f zx = f xz for all ( x , y , z ) in the domain of f . Thus,

f xzx = f zxx =2x 2

.

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Exercises

1 Find f x and f y for f (x , y ) =tan(3 x y )+x y

.

2 Find g x and g y for g (x , y ) =x y .

3 Find h x and h y for h (x , y ) =x y

x 2 y .4 Find f x , f y and f z for f (x , y , z ) =log x ( y )xz

2 .

5 Solve for 2 f x z

if f (x , y , z ) =x 2 3 y

z .

6 Evaluate f yxz (1,2,0) for f (x , y , z ) =9x 2z e

2x y .

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25 Partial Derivatives

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