UNIT 2

PARTIAL DIFFERENTIATION 1

Let

Then partial derivative of z with respect to x is obtained by differentiating z with respect to x treating y as constant and is denoted as

Then partial derivative of z with respect to y is obtained by differentiating z with respect to y treating x as constant and is denoted as

Partial derivative of higher order:

When we differentiate a function depend on more than one independent variable, we differentiate it with respect to one variable keeping other as constant.

A second order partial derivative means differentiating twice

In general are also function of x and y and so these can be further partially differentiated with respect to x and y.

In general

Notation:

Generalization: If

Then the partial derivative of z with respect to is obtained by differentiating z with respect to treating all the other variables as constant and is denoted by

Example1:If . Then prove that

Given

Partially differentiating z with respect to x keeping y as constant

Again partially differentiating given z with respect to y keeping x as constant

On b.eq(i) +a.eq(ii) we get

Hence proved.

Example2 : If

Show that

Given

Partially differentiating z with respect to x keeping y as constant

Again partially differentiating z with respect to x keeping y as constant

Partially differentiating z with respect to y keeping x as constant

Again partially differentiating z with respect to y keeping x as constant

From eq(i) and eq(ii) we conclude that

Example3 : Find the value of n so that the equation

Satisfies the relation

Given

Partially differentiating V with respect to r keeping as constant

Again partially differentiating given V with respect to keeping r as constant

Now, we are taking the given relation

Substituting values using eq(i) and eq(ii)

On solving we get

Example 4 : If then show that when

Given

Taking log on both side we get

Partially differentiating with respect to x we get

…..(i)

Similarly partially differentiating with respect y we get

……(ii)

LHS

Substituting value from (ii)

Again substituting value from (i) we get

.()

When

=RHS

Hence proved

Example5 :If

Then show that

Given

Partially differentiating u with respect to x keeping y and z as constant

Similarly paritially differentiating u with respect to y keeping x and z as constant

…….(ii)

……..(iii)

LHS:

Hence proved

Homogenous Function: A function of the form

In this every term if of degree n, so it is called as homogenous function of degree n.

Rewriting the above as

=

=

Thus every function which can be expressed as above form is called a homogenous function of degree n in x and y.

Generalization: A function is an homogenous equation of degree n in if it can be expressed as

Euler’s Theorem:

If u be an homogenous function of degree n in x and y, then

Proof: Given u is an homogenous function of degree n in x and y. So it can be rewrite as

……(i)

Partially differentiating u with respect to x we get

Again partially differentiating u with respect to y we get

Therefore

Thus

Hence proved

Corollary: If u is a homogenous function of degree n in x and y then

As we know that by Euler’s theorem

……(i)

Partially differentiating (i) with respect to x we get

…..(ii)

Partially differentiating (i) with respect to y we get

…..(iii)

Multiplying x by (ii) and y by (iii) then on adding we get

by using (i)

Thus

Note: We can directly use the Euler’s theorem and its corollary to solve the problems.

Example1 Show that

Given

Therefore f(x,y,z) is an homogenous equation of degree 2 in x, y and z

Example2 If

Let

Thus u is an homogenous function of degree 2 in x and y

Therefore by Euler’s theorem

substituting the value of u

Hence proved

Example3If , find the value of

Given

Thus u is an homogenous function of degree 6 in x ,y and z

Therefore by Euler’s theorem

Example4 If

Given

Thus u is an homogenous equation of degree -1 in x and y

Therefore by Euler’s theorem

Chain rule for two sets of independent variables

If u=u(x,y) and the two independent variables x,y are each a function of two new independent variables s,t then we want relations between their partial derivatives.

1.when u =u(x,y),

Conversely, when u= u(s,t).

Then when s =s(x,y) and t =t(x,y).

Example:1

Solve the following using chain Rule

Y=

Solution:

Then ,

Substituting U = in the above we have

Example 2:

There really isn’t all that much to do here other than using the formula

So, technically we’ve computed the derivative .However ,we should probably go ahead and substitute in for x and y as well doing this gives

Right actually have been easier to just substitute in for x and y in the original function and just compute the derivative.

Reference Books:

1. A text book of Applied Mathematics Volume I and II by J.N. Wartikar and P.N. Wartikar

2. Higher Engineering Mathematics by Dr. B. S. Grewal

3. Advanced Engineering Mathematics by H. K. Dass

4. Advanced Engineering Mathematics by Erwins Kreyszig