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 - xy⁵)

Solution- 3x²y² - y⁵, 2x³y – 5xy⁴,

= = 6xy²

= 2x³ - 20xy³

= = 6x²y – 5y⁴

= = 6x²y - 5y⁴

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

f( kx, ky) = kⁿf(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 = xⁿ f(y/x) ----------------(1)

Partially differentiate equation (1) with respect to x,

= n f(y/x) + xⁿ f’(y/x).()

Now multiplying by x on both sides, we get

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

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

= xⁿ f’(y/x).

Now multiplying by y on both sides,

y = xⁿ f’(y/x). ---------------(3)

By adding equation (2) and (3),

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

3x²y⁴3t² + 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 = x² + 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