UNIT 2

Partial Differentiation

2.1 Functions of several variables

There are two types of quantities whose value depend on a single variable and another whose value depend on more than a single variable.

In other words : A symbol z which has a definite value for each pair of values of x and y is called a function of two independent variables x and y, denoted by

Example: velocity depends on distance and time.

Volume of cylinder depends on height and radius of cylinder.

Consider a function z which has a definite value for the independent variables

Then it is called as function of several variables and is denoted by

A function whose value dependent on several independent variables is called as function of several variables.

2.2 First and Higher order derivatives

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

2.3 Euler’s theorem

A polynomial in x & y is said to be Homogeneous expression in x & y of degree n. If the degree of each term in the expression is same & equal to n.

e.g.

is a homogeneous function of degree 3.

To find the degree of homogeneous expression f(x, y).

1. Consider
2. Put . Then if we get .

Then the degree of is n.

Ex.

Consider

Put

.

Thus degree of f(x, y) is

Note that

If be a homogeneous function of degree n then z can be written as

Differentiation of Implicit function

Suppose that we cannot find y explicitly as a function of x. But only implicitly through the relation f(x, y) = 0.

Then we find

Since

diff. P. w.r.t. x we get

i.e.

Similarly,

It f (x, y, z) = 0 then z is called implicit function of x, y. Then in this case we get

Ex.

Find if

Ex. Find . If , &

Ex. If , where

Find

Ex. If

Then find

Eulers Theorem on Homogeneous functions:

Statement:

If be a homogeneous function of degree n in x & y then,

Deductions from Eulers theorem

1. If be a homogeneous function of degree n in x & y then,

.

2.     If be a homogeneous functions of degree n in x & y and also then,

And

Where

Ex.

If , find the value of

Ex.

If then find the value of

Ex. If then prove

That

Ex. If the prove that

Ex. If  then show

That

2.4 Chain rule and total differential coefficient

In calculus, the general Leibniz rule named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). It states that if {\displaystyle f}f  and {\displaystyle g}g are {\displaystyle n}n-times differentiable functions, then the product {\displaystyle fg}fg is also {\displaystyle n}n-times differentiable and its {\displaystyle n}nth derivative is given by

(fg)n = k=0n  (nk) f(n-k).g(k)

{\displaystyle (f\circ g)'=(f'\circ g)\cdot g'.}Alternatively, by letting F = f ∘ g (equiv., F(x) = f(g(x)) for all x), one can also write the chain rule in Lagrange's notation, as follows:

F’(X)= f’(g(x)).g’(x)

{\displaystyle F'(x)=f'(g(x))g'(x).}The chain rule may also be rewritten in Leibniz's notation in the following way. If a variable z depends on the variable y, which itself depends on the variable x (i.e., y and z are dependent variables), then z, via the intermediate variable of y, depends on x as well. In which case, the chain rule states that:

{\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}.} =

The Chain Rule Formula is as follows – 

= .

Solved Examples

Example 1: Differentiate y = cos x2

Solution:

Given,

y = cos x2

Let u = x2, so that y = cos u

Therefore:  =2x

= -sin u

And so, the chain rule says:

=  .

= -sin u × 2x

= -2x sin x2

Example 2: 

Differentiate f(x)=(1+x2)5.

Solution: 

Using the Chain rule,

=

Let us take y = u5 and u = 1+x2

Then =  (u5) = 5u4

=   (1 + x2 )= 2x.  

= 5u4⋅2x = 5(1+x2)4⋅2x

= 10x(1+x4)

Implicit differentiation:

A function can be explicit or implicit:

Explicit: "y = some function of x". When we know x we can calculate y directly.

Implicit: "some function of y and x equals something else". Knowing x does not lead directly to y.

How to do Implicit Differentiation

• Differentiate with respect to x
• Collect all the  on one side
• Solve for 

Example 1: x2 + y2 = r2

Differentiate with respect to x:

(x2) + (y2) = (r2)

Let's solve each term:

Use the Power Rule: (x2) = 2x

Use the Chain Rule (explained below): (y2) = 2y

r2 is a constant, so its derivative is 0:d/dx(r2) = 0

Which gives us:

2x + 2y= 0

Collect all the  on one side

Y = −x

Solve for :

 =- 

The Chain Rule Using   :

Let's look more closely at how  (y2) becomes 2y

The Chain Rule says:

=.

Substitute in u = y2:

(y2) = (y2) .

And then:

(y2) = 2y

Basically, all we did was differentiate with respect to y and multiply by dy/dx

Example 2:

The following problems require the use of implicit differentiation. Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions y written EXPLICITLY as functions of x. For example, if

 ,

Then the derivative of y is

 .

However, some functions y is written IMPLICITLY as functions of x. A familiar example of this is the equation

x2 + y2 = 25,

Which represents a circle of radius five cantered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .

How could we find the derivative of y in this instance? One way is to first write y explicitly as a function of x. Thus,

x2 + y2 = 25,

y2 = 25 - x2,

And

 ,

Where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point

(3, -4) lies on the bottom semi-circle given by

 ,

The derivative of y is

 ,

i.e.,

 .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

 .

Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))2 , can be found using the chain rule :

 .

Since y symbolically represents a function of x, the derivative of y2 can be found in the same fashion :

 .

Now begin with

x2 + y2 = 25 .

Differentiate both sides of the equation, getting

D ( x2 + y2 ) = D ( 25 ) ,

D ( x2 ) + D ( y2 ) = D ( 25 ) ,

And

2x + 2 y y' = 0 ,

So that

2 y y' = - 2x ,

And

 ,

i.e.,

 .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

 .

2.5 Jaccobians

Jacobians, Errors and Approximations, maxima and minima

Jacobians

If u and v be continuous and differentiable functions of two other independent variables x and y such as

, then we define the determine

as Jacobian of u, v with respect to x, y

Similarly ,

JJ’ = 1

Actually Jacobins are functional determines

Ex.

1.   Calculate
2. If
3. If

ST

4.      find

5.     If and , find

6.    

7.     If  

8.     If , ,

JJ1 = 1

If ,

JJ1=1

Jacobian of composite function (chain rule)

Then

Ex.

1. If

Where

2.     If   and

Find

3.     If

Find

Jacobian of Implicit function

Let u1, u2 be implicit functions of x1, x2 connected by f1, f2 such there

,

Then

Similarly,

Ex.

If

If

Find

Partial derivative of implicit functions

Consider four variables u, v, x, y related by implicit function.

,

Then

Ex.

If and

Find

If and

Find

Find

If

Find

2.6 Taylor’s & Maclaurin’s series for two variables

Taylor series:

The taylor’s series can be represented as the following

  (x-a)n

Example 1:

Find the taylor series for the following:

= <1

(x/10)<1 and (x/10) > -1

Therefore radius of convergence is (-10,10)

  ROC =10

Example 2:

f(n)5 =

Here the ROC is 4

Maclaurian series:

  (x)n

Example:

f(x)=

= f(0)+f’(0)x+ x2 + x3 +......

= 1+x+x2 +x3 + .....

=

2.7 Maxima & Minima of functions of two variables

As we know that the value of a function at maximum point is called maximum value of a function. Similarly, the value of a function at minimum point is called minimum value of a function.

The maxima and minima of a function is an extreme biggest and extreme smallest point of a function in a given range (interval) or entire region. Pierre de Fermat was the first mathematician to discover general method for calculating maxima and minima of a function. The maxima and minima are complementing of each other.

Maxima and Minima of a function of one variables

If  f(x) is a single valued function defined in a region R then

Maxima is a maximum point if and only if

Minima is a minimum point if and only if

Maxima and Minima of a function of two independent variables

Let be a defined function of two independent variables.

Then the point is said to be a maximum point of if

Or   =

For all positive and negative values of h and k.

Similarly the point is said to be a minimum point of if

Or   =

For all positive and negative values of h and k.

Saddle point: Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.

A point is a saddle point of a function of two variables if

At the point.

Stationary Value

The value is said to be a stationary value of if

i.e. the function is a stationary at (a , b).

Rule to find the maximum and minimum values of

1. Calculate.
2. Form and solve , we get the value of x and y let it be pairs of values
3. Calculate the following values :

4.     (a) If

(b) If

(c) If

(d) If

Example1 Find out the maxima and minima of the function

Given     …(i)

Partially differentiating (i) with respect to x we get

    ….(ii)

Partially differentiating (i) with respect to y we get

    ….(iii)

Now, form the equations

Using (ii) and (iii) we get

          using above two equations

Squaring both side we get

Or 

This show that

Also we get

Thus we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0)  we get

So, the point (0,0) is a saddle point.

At point we get

So the point is the minimum point where

In case 

So the point is the maximum point where

Example2 Find the maximum and minimum point of the function

Partially differentiating given equation with respect to and x and y then equate them to zero                 

On solving above we get 

Also

Thus we get the pair of values (0,0), (,0) and (0,

Now, we calculate

At the point (0,0)

So function has saddle point at (0,0).

At the point (

So the function has maxima at this point (.

At the point (0,

So the function has minima at this point (0,.

At the point (

So the function has an saddle point at (

Example3 Find the maximum and minimum value of

Let

Partially differentiating given function with respect to x and y and equate it to zero

  ..(i)

    ..(ii)

On solving (i) and (ii) we get

Thus pair of values are

Now, we calculate

At the point (0,0)

So further investigation is required

On the x  axis y  = 0  , f(x,0)=0

On the line y=x,

At the point

So that the given function has maximum value at

Therefore maximum value of given function

At the point

So that the given function has minimum value at

Therefore minimum value of the given function

2.8Langrage’s method of undetermined multipliers

Generally to calculate the stationary value of a function with some relation by converting the given function into the least possible independent variables and then solve them. When this method fail we use Lagrange’s method.

This method is used to calculate the stationary value of a function of several variables which are all not independent but are connected by some relation.

Let  be the function in the variable x, y and z which is connected by the relation

Rule: a) Form the equation

Where is a parameter.

b) Form the equation using partial differentiation is

(We always try to eliminate)

c) Solve the all above equation with the given relation

These give the value of

These value obtained when substituted in the given function will give the stationary value of the function.

Example1 Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.                                    

Let first number be x, second be y and third be z.

According to the question

Let the  given function  be  f

And the relation

By Lagrange’s Method

   ….(i)

Partially differentiating (i) with respect to x,y and z and  equate  them  to zero

   ….(ii)

   ….(iii)

   ….(iv)

From (ii),(iii)  and (iv) we get

On solving

Putting it in given relation we get

Or 

Or

Thus the first number is 4 second is 8 and third is 12

Example2 The temperature T at any point   in space is  .Find  the  highest temperature on the surface of the unit sphere.

Given function is  

On the surface of unit sphere given     [ is an equation of unit sphere in 3   dimensional space]

By Lagrange’s Method

     ….(i)

Partially differentiating (i) with respect to  x, y and z and equate them to zero

or           …(ii)

or           …(iii)

          …(iv)

Dividing (ii) and (ii) by (iv) we get

Using given relation 

Or       

Or        

So that

Or 

Thus points are

The maximum temperature is

Example3  If ,Find the value of x and y for  which is maximum.

Given function is

And relation is

By Lagrange’s Method

[]     ..(i)

Partially differentiating (i) with respect to  x, y and z and equate them to zero

Or       …(ii)

Or       …(iii)

Or       …(iv)

On solving (ii),(iii) and (iv) we  get

Using the given relation  we get

So that

Thus the point for the maximum value of the given function is

Example4  Find  the points on the surface nearest to the origin.

Let be any point on the surface, then its distance from the origin is

Thus the given equation will be

And relation is

By Lagrange’s Method

     ….(i)

Partially differentiating (i) with respect to  x, y and z and equate them to zero

Or        …(ii)

Or        …(iii)

Or

Or

On solving equation  (ii) by (iii) we get

And     

On subtracting we  get

Putting in above

Or 

Thus

Using the given relation we  get

= 0.0 +1=1

Or 

Thus point on the surface nearest to the origin is