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