Presentation on Partial Differentiation and its Application

Partial DifferentiationDefinition :-

A partial differential equation is an equation involving a function of two or more variables and some of its partial derivatives. Therefore a partial differential equation contains one dependent variable and more than one independent variable. Here z will be taken as the dependent variable and x and y the independent variable so that .

We will use the following standard notations to denote the partial derivatives.

yxfz , .

,, qyzp

xz

t

yzs

yxzr

xz

2

22

2

2

,,

Partial Derivatives of First Orderf (x, y) is a function of two variables. The first order partial derivative of f with respect to x at a point (x, y) is

0

( , ) ( , )limh

f f x h y f x yx h

f (x, y) is a function of two variables. The first partial derivative of f with respect to y at a point (x, y) is

0

( , ) ( , )limk

f f x y k f x yy k

∂f/∂x and ∂f/∂y are called First Order Partial

Derivatives of f .

Example

Partial Derivatives of First Order ExampleCompute the first order partial derivatives

2 3( , ) 3 2 .f x y x y y 22 33 6 yxfxyf yx

Compute the first order partial derivatives Example

432),( yxyxf

3 2 yx ff

Partial Derivatives of Higher Order

if z(x, y) is a function of two variables. Then ∂z/∂x and ∂z/∂y are also functions of two variables and their partials can be taken. Hence we can differentiate with respect to x and y again and find ,

Definition :-

1. 2zxx

2 fxx

2zx2

2 fx2 fxx

Take the partial withrespect to x, and thenwith respect to x again.

2.2zyx

2 fyx

fxy

Take the partial withrespect to x, and thenwith respect to y.

3.2zxy

2 fxy

fyx

Take the partial withrespect to y, and thenwith respect to x.

4.2zyy

2 fyy

2zy2

2 fy2 fyy

Take the partial with respect to y, and then with respect to y again.

Example

Partial Derivatives of Higher Order Example

Example

Find the second-order partial derivatives of the function2( , ) 3 lnf x y x y x y

yxf

yxf

yxfyf yxxyyyxx

16166 2

6 lnxf xy y 2 13yf x xy

Find the second-order partial derivatives of the function2a. ( , ) 3 2f x y x y

yff yx 43

0f 0f 4f 0f yxxyyyxx

Find the higher-order partial derivatives of the function2

( , ) xyf x y eExample

2 22xy xyxf e y e

x

2 2

2xy xyyf e xye

y

24 xyxxf y e 2 22 1xy

yxf ye xy

2xyxy2xyxy xy1ye2xye2yye2f

222

2xyxyxyyy xy21xe2xy2eyex2f

222

Homogeneous Functions A function f(x,y) is said to be homogeneous function of degree n if it can be expressed as

xyxn

yxynOR

Euler’s Theorem on Homogeneous Functions if f is a homogeneous function of x and y of degree n then

nfyfy

xfx

if f is a homogeneous function of degree n, then

fnny

fyyxfxy

xfx )1(2 2

22

2

2

22

Euler’s Theorem Example

The Chain Rule

Suppose z = f(x, y) is a differentiable function of x and y, where x = g(s, t) and y = h(s, t) are differentiable functions of s and t.Then,

z z x z y z z x z ys x s y s t x t y t

If u=f(x,y), where x=g(t) and y=h(t) then we can express u as a function of t alone by substituting the value of x and y in f(x,y).Now to find du/dt without actually substituting the values of x and y in f(x,y), we establish the following Chain rule :-

dtdy

yu

dtdx

xu

dtdu

The Chain Rule ExampleIf z = ex sin y, where x = st2 and y = s2t, find ∂z/∂s and ∂z/∂t.

Example

Applying the Chain Rule, we get the following results.

2 2

2

2 2 2

( sin )( ) ( cos )(2 )

sin( ) 2 cos( )

x x

st st

z z x z ys x s y s

e y t e y st

t e s t ste s t

2 2

2

2 2 2

( sin )(2 ) ( cos )( )

2 sin( ) cos( )

x x

st st

z z x z yt x t y t

e y st e y s

ste s t s e s t

And

JACOBIANSIf x and y are functions of two independent variables u and v ,then the determinant Is called Jacobian of x , y

w.r.t. u , v

and is denoted by

Properties of Jacobians The Jacobian matrix is the Inverse matrix of i.e.,1.

2. The Multiplication of jacobian matrix and is = 1.

. =1.

Example of Jacobians

Taylor’s theorem for a function of two variables

Now let f(x , y ) be the function of two independent variables x and y. If y is kept constant then Taylor’s theorem for a function of a single variable x

Putting x=a and y=b , we have

Taylor’s theorem for a function of two variables

Putting a+h=x and b+k=y so that h=x-a and k=y-b , we have

Maclaurin’s Theorem of two variablePutting a=0 ,b=0 , we have

Taylor’s theorem for a function of two variables Example

Example

Maxima And Minima for a function of two variables

• The Function f(x,y) is maximum at (x,y) if for all small positive or negative values of h and k; we have

• f(x+h , y+k) – f(x,y) < 0

• Similarly f(x,y) is minimum at (x,y) if for all small positive or negative values of h and k, we have

• f(x+h , y+k) – f(x,y) > 0

• Thus ,from the defination of maximum of f(x,y) at (x,y) we note that f(x+h , y+k) – f(x,y) preserves the same sign for a maximum it is negative and for a minimum it is positive

Working rule for Maxima And Minima for a function of two variables

• Working rule to find maximum and minimum values of a function f(x,y)

• (1) find ∂f/∂x and ∂f/∂y• (2) a necessary condition for maximum or minimum value is

∂f/∂x=0 , ∂f/∂y=0• solve simultaneous equations ∂f/∂x=0 , ∂f/∂y=0• Let (a₁,b₁) , (a₂,b₂) … be the solutions of these equations.• Find ∂²f/∂x²=r , • ∂²f/∂x ∂y=s , • ∂²f/∂y²=t• (3) if rt-s²>0 and r>0 at one or more points then those are

the points of minima.

Working rule for Maxima And Minima for a function of two variables

• (4) if rt-s²>0 or r<0 at one or more points then those points are the points of maxima.

• (5) if rt-s²<0 ,then there are no maximum or minimum at these points. Such points are called saddle points.

• (6) if rt-s²=0 nothing can be said about the maxima or minima .it requires further investigation.

Example for Maxima And Minima for a function of two variables

Example

Discuss the maxima and minima of f(x,y)= xy+27(1/x + 1/y)

∂f/∂x=y-(27/x²) , ∂f/∂y=x-(27/y²)For max. or min ,values we have ∂f/∂x=0 , ∂f/∂y=0.y-(27/x²)=0…(1)x-(27/y²)=0…(2)Giving x=y=3

²f/∂x²=r =54/x³ ∂²f/∂x ∂y=s=1 , ∂²f/∂y²=t=27/y³r(3,3)=3s(3,3)=1t(3,3)=3rt-s²=9-1=8>o , since r,t are both >0

We get minimum value at x=y=3 which is 27.

Lagrange’s MethodA method to find the local minimum and maximum of a function with two variables subject to conditions or constraints on the variables involved.Suppose that, subject to the constraint g(x,y)=0, the function z=f(x,y) has a local maximum or a local minimum at the point

. Form the function

, , , ,F x y f x y g x y Then there is a value of such that is a solution of the system of equations

0 0( , , )x y

0 1

0 2

, 0 3

F f gx x xF f gy y yF g x y

provided all the partial derivatives exists.

Working Rule for Lagrange’s MethodStep 1: Write the function to be maximized (or minimized) and the

constraint in the form: Find the maximum (or minimum) value of

subject to the constraintStep 2: Construct the function F:

,z f x y

, 0g x y

, , , ,F x y f x y g x y

Step 3: Set up the system of equations

0 1

0 2

, 0 3

FxFyF g x y

Working Rule for Lagrange’s MethodStep 4: Solve the system of equations for x, y and .Step 5: Test the solution to determine maximum or

minimum point.

0 0( , , )x y

Find D* = Fxx . Fyy - (Fxy)2

If D* 0 Fxx 0 maximum pointFxx 0 minimum point

D* 0 Test is inconclusiveStep 6: Evaluate at each solution

found in Step 5. ,z f x y 0 0( , , )x y

Example for Lagrange’s Method

Find the minimum off(x,y) = 5x2 + 6y2 - xy

subject to the constraintx+2y = 24

Solution:F(x,y, ) = 5x2 + 6y2 - xy + (x + 2y - 24)Fx = F = 10x - y + ; Fxx = 10

xFy = F = 12y - x + 2 ; Fyy = 12

yF = F = x + 2y - 24 ; Fxy = -1

Example

The critical point,10x - y + = 012y - x + 2= 0x + 2y - 24= 0

The solution of the system is x = 6, y = 9, = -51

D*=(10)(12)-(-1)2=119>0Fxx = 10>0

We find that f(x,y) has a local minimum at (6,9).

f(x,y) = 5(6)2+6(9)2-6(9)= 720

Differentiation Under Integral SignDifferentiation under the integral sign is an operation is used to evaluate certain integrals. Under fairly loose conditions on the function being integrated, differentiation under the integral sign allows one to interchange the order of integration and differentiation. In its simplest form, called the Leibniz integral rule, differentiation under the integral sign makes the following equation valid under light assumptions on f :

This simpler statement is known as Leibniz integral rule.Examples

Example for Differentiation Under Integral Sign

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