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M1


Unit – 2


Partial Differentiation


Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:

=

 

Now the partial derivative of f with respect to f can be written as and defined as follows:

=

Note: a. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant.

b. We apply all differentiation rules.

 


Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.

These are second order four partial derivatives:

(a)     =            

(b)  =                 

(c)   =                 

(d)     =

b and c are known as mixed partial derivatives.

Similarly we can find the other higher order derivatives.

 

Now we will understand the partial derivative by some examples:

Example1- Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

Solution:   To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 9x² - 0 + 2y – 8 + 0 – 0

= 9x² + 2y – 8

Similarly partial derivative of f(x,y) with respect to y is:

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 0 – 10y + 2x – 0 + 4 – 0

= 2x – 10y +4.

 

Example2 - Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

Solution-   :   To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= (y² + 50) cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= 2xy cos(y²x + 5x – 8)

 

Example3-   Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy)

Solution-    3x²y² - y,      2x³y – 5xy,

  = = 6xy²

  = 2x³ - 20xy³

= = 6x²y – 5y    

= = 6x²y -   5y

 


(a) Homegeneous function -  A function f(x,y) is said to be homogeneous of degree n if,

f( kx, ky) =  kf(x, y)

Here, the power of k is called the degree of homogeneity.

 

(b) Euler’s theorem (proof)-

Statement – if u = f(x, y) be a homogeneous function in x and y of degree n , then

x + y = nu

Proof:   Here u is a homogeneous function of degree n,

u = x f(y/x)        ----------------(1)

Partially differentiate equation (1) with respect to x,

= n f(y/x) + x f’(y/x).()

Now multiplying by x on both sides, we get

x = n f(y/x) + x f’(y/x).()  ---------- (2)

 

Again partially differentiate equation (1) with respect to y,

= x f’(y/x).

Now multiplying by y on both sides,

y = x f’(y/x).      ---------------(3)

By adding equation (2) and (3),

xy = n f(y/x) + + x f’(y/x).()  + x f’(y/x).

xy         =  n f(y/x)

Here u = f( x, y) is homogeneous function, then -  u = f(y/x)

Put the value of u in equation (4),

xy   =  nu

Which is the Euler’s theorem.

Lets understand Eulers’s theorem by some examples:

Example1-  If u = x²(y-x) + y²(x-y), then show that     -2 (x – y)²

Solution -  here,  u = x²(y-x) + y²(x-y)

u = x²y - x³ + xy² - y³,

Now differentiate u partially with respect to x and y respectively,

 =  2xy – 3x² + y²        --------- (1)

= x² + 2xy – 3y²          ---------- (2)

Now adding equation (1) and (2), we get

= -2x² - 2y² + 4xy

= -2 (x² + y² - 2xy)

= -2 (x – y)²   

Example2-  If u = xy + sin(xy), show that    =  .

Solution –            u = xy + sin(xy)

  =   y+ ycos(xy)

=  x+ xcos(xy)

x (- sin(xy).(y)) + cos(xy)

= 1 – xysin(xy) + cos(xy)     -------------- (1)

1 + cos(xy) + y(-sin(xy) x)

= 1 – xy sin(xy) + cos(xy)      -----------------(2)

From equation (1) and (2),

  = 

 


When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant.

Let the function,     u = f( x, y),  such that    x = g(t)  ,     y = h(t)

Then we can write,

=

=

This is the total derivative of u with respect to t.

Example1:   let q = 4x + 3y      and    x = t³ + t² + 1    , y = t³ - t² - t

Then find  .

Solution:            . =

Where, f1 =         ,  f2 = 

 

In this example f1 = 4    ,      f2 = 3

Also,              3t² + 2t    ,                      

4(3t² + 2t) + 3(

=  21t² + 2t – 3

 

Example2:    Find      if u = x³y⁴    where   x = t³     and    y = t².

Solution:    as we know that by definition,     =

3x²y3t² + 4x³y³2t = 17t¹⁶.

 


If w = f (x, y) has continuous partial variables fx and fy and if x = x (t), y = y (t) are

Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a

Differentiable function of t.

In this case, we get,

fx (x (t), y (t)) x’ (t) +  fy (x(t), y (t)) y’ (t).

 

Example1:  if w = x² + y – z + sint and  x + y = t, find

 

(a)    y,z

 

(b)   t, z    

Solution: With x, y, z independent, we have

t = x + y, w = + y - z + sin (x + y).

Therefore,

y,z = 2x + cos(x+y)(x+y)

=  2x + cos (x + y)

With x, t, z independent, we have

Y = t-x,   w=  x² + (t-x) + sin t

Thus     t, z   =  2x - 1

 


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