Partial differentiation B tech

Definition :-
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 , .
,, q
y
z
p
x
z






t
y
z
s
yx
z
r
x
z









2
22
2
2
,,

f (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
( , ) ( , )
lim
h
f f x h y f x y
x 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
( , ) ( , )
lim
k
f f x y k f x y
y k
  


∂f/∂x and ∂f/∂y are called First Order Partial Derivatives of f .

Example Compute the first order partial derivatives
2 3
( , ) 3 2 .f x y x y y  
22
336 yxfxyf yx 
Compute the first order partial derivativesExample
432),(  yxyxf
32  yx ff

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.
2
z
xx

2
f
xx

2
z
x2

2
f
x2
 fxx
Take the partial with
respect to x, and then
with respect to x again.
2.
2
z
yx

2
f
yx
 fxy
respect to x, and then
with respect to y.
3.
2
z
xy

2
f
xy
 fyx
respect to y, and then
with respect to x.
4.
2
z
yy

2
f
yy

2
z
y2

2
f
y2
 fyy
respect to y, and then
with respect to y again.

Example
Example
Find the second-order partial derivatives of the function
2
( , ) 3 lnf x y x y x y 
y
xf
y
xf
y
x
fyf yxxyyyxx
1
6
1
66 2

6 lnxf xy y  2 1
3yf x x
y
 
   
 
Find the second-order partial derivatives of the function
2
a. ( , ) 3 2f x y x y 
yff yx 43 
0f0f4f0f yxxyyyxx 

Find the higher-order partial derivatives of the function
2
( , ) xy
f x y e
Example
 2 2
2xy xy
xf e y e
x

 
  2 2
2xy xy
yf e xye
y

 

2
4 xy
xxf y e  
2
2
2 1xy
yxf ye xy 
   2xyxy2xy
xy xy1ye2xye2yye2f
222

    2xyxyxy
yy xy21xe2xy2eyex2f
222


A function f(x,y) is said to be homogeneous function of degree n if it
can be expressed as






x
y
xn
 





y
x
yn
OR
if f is a homogeneous function of x and y of degree n then
nf
y
f
y
x
f
x 





if f is a homogeneous function of degree n, then
fnn
y
f
y
yx
f
xy
x
f
x )1(2 2
2
2
2
2
2
2










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 y
s 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 :-
dt
dy
y
u
dt
dx
x
u
dt
du








If 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 y
s 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 y
t x t y t
e y st e y s
ste s t s e s t
And

If 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

The Jacobian matrix is the Inverse matrix of i.e.,1.
2. The Multiplication of jacobian matrix and is = 1.
. =1.

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

Putting a+h=x and b+k=y so that h=x-a and k=y-b , we have
Maclaurin’s Theorem of two variable
Putting a=0 ,b=0 , we have

• 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 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.

• (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
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/y³
r(3,3)=3
s(3,3)=1
t(3,3)=3
rt-s²=9-1=8>o , since r,t are both >0
We get minimum value at x=y=3 which is 27.

A 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 g
x x x
F f g
y y y
F
g x y



  
  
  
  
  
  

 

L
L
L
provided all the partial derivatives exists.

Step 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 constraint
Step 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
F
x
F
y
F
g x y








 

L
L
L

Step 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 point
Fxx  0  minimum point
D*  0  Test is inconclusive
Step 6: Evaluate at each solution
found in Step 5.
 ,z f x y 0 0( , , )x y 

Find the minimum of
f(x,y) = 5x2 + 6y2 - xy
subject to the constraint
x+2y = 24
Solution:
F(x,y, ) = 5x2 + 6y2 - xy + (x + 2y - 24)
Fx = F = 10x - y +  ; Fxx = 10
x
Fy = F = 12y - x + 2 ; Fyy = 12
y
F = F = x + 2y - 24 ; Fxy = -1

Example

The critical point,
10x - y +  = 0
12y - x + 2= 0
x + 2y - 24= 0
The solution of the system is x = 6, y = 9,  = -51
D*=(10)(12)-(-1)2=119>0
Fxx = 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 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

Raj Kumar (1611251011)
Abhinav (1611254004)
Jasbeer (1611254006)
Baljeet (1611254007)

Partial differentiation B tech

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