Unit – 1

Function of one variable

1.1 Function of one variable: Successive Differentiation

It is the process of differentiating the given function simultaneously many times and  the result   obtained are  called successive  derivative.

Let be a differentiable function.

First derivative is denoted by

Second derivative is  

Third derivative is

Similarly the nth derivative is

Successive differentiation-

The successive differential coefficients of y are denoted as follows-

……………….

The differential coefficient is-

Other notations to denote n’th differential coefficients-

The process of applying differentiation again and again is called successive differentiation.

nth derivative of standard functions-

1. nth derivative of –

Suppose y = ,

Differentiate with respect to x successively, we get

For n times differentiation, we get-

So we can say that its n’th derivative will be

2. nth derivative of log(ax + b)-

Suppose y = log (ax + b)

Differentiate with respect to x successively, we get

(-2)

For n times differentiation, we get-

(-2)…………….(-n + 1)

=  …………….(n - 1)

= (n - 1)

So we can say that its n’th derivative will be

3. nth derivative of

Suppose y =

Differentiate with respect to x successively, we get

For n times differentiation, we get-

So we can say that its n’th derivative will be

4.  nth derivative of sin(ax + b)-

Suppose y = sin(ax + b)

Differentiate with respect to x successively, we get

For n times differentiation, we get-

So we can say that its n’th derivative will be

5. Nth derivative of

Let us consider the functions-

To rewrite this in the form of sin, put-

Diff. Again w.r.t.x, we get-

Substitute for a and b, we get-

…………………………………………………………………………

Similarly we get-

And

Similarly we can find the ‘n’ derivatives of such functions-

Standard results-

1.

2.

3.

4.

5.

6.

7.

Example:  If y = l , then show that-

Sol. We have,

y =

Differentiate y with respect to x, we get

=

Again diff. (n – 1) times w.r .t x , we get-

Order differential equation

N’th order differential equations can be solved as below-

Example: Find the derivative of

Sol. Here we have-

Suppose,  y =

First derivative-

Here ,

Let x =

So that

Which is the n’th derivative of the given function.

Example: Find cos x cos 2x cos 3x.

Sol.

So that-

n’th derivative-

derivatives of algebraic functions, derivatives of function belongs to polar form

Example: Find the derivative of the following function-

Sol.  Partial fraction of the function y after splitting-

Suppose x – 1 = z, then

=

=

=

Here we can find its n’th derivative-

Example: Find derivative of the given function:

y =

Sol.   We are given-

y =

Factorize the denominator-

y =

=

derivative will be-

Which is the derivative of the given function.

Example: Find if

Sol.

Here we have-

At x = 0,

When n is odd-

When n is even

Example: Find the nth derivative of sin3 x

Sol: we know that sin 3x= 3sin x 4sin3 x =  sin3x  = 

Differentiate   n    times w.r.t x,

( sin3 x) = (3 sinx- sin3x)

=  (  -3n. Sin(  3x+ nπ/2)  +   3 sin (x+  nπ/2)) nϵz

Example: Find the nth derivative of sin 5x. Sin 3x.?

Sol: let y =   sin 5x.sin 3x= ( sin 5x.sin 3x)  

⇒y= (  cos 2x -  cos8x)

⇒ y= ( cos 2x- cos8x )

Differentiate n times w.r.t x,

Yn  = ( cos 2x -  cos8x )

⇒ yn = ( 2 n (cos( 2x+ nπ/2)-  8n .cos (8x + nπ/2)) nϵz.

Successive n th derivative of nth elementary function ie., exponential

Example: If  y  = ae n x +  be –nx ,    then show that   y2= n2y

Sol: Y= aenx + be-nx

y 1 =   a.n.enx  - b.n.e-nx

y2 = an2 enx – bn2 e-nx  = n2 (ae nx+ be –nx)

y2= n2y. 

Example: If  y= e-kx/2(a cosnx+ b sinnx) then show that.,y2+ ky1+(n2+ k2/4)y =0

Sol: y= e-kx/2(a cosnx+ b sinnx)

Differentiating  w.r.to. x.,

Y1 = e-kx/2( -an sin nx + bn cos nx) - k/2.y

Y1+ k/2.y = ne-kx/2 ( -an sin nx + bn cos nx)    (1)

Differentiating w.r.to x.,

Y2+ k/2.y1 = ne-kx/2  (-k/2) ( -an sin nx + bn cos nx) + n e-kx/2(-an  cosnx- bn  sinnx).

= -(k/2) (y1+ k/2 y)- n2 y = - (k/2 y1)- ( k2/4)y- n2y.

y2 + ky1 +(n2+ k2/4)y = 0.

Key takeaways-

1.

2.

3.

4.

5.

6.

7.

1.2 Leibnitz Theorem

Statements of Leibnitz’s Theorem-

If u and v are the function of x  such  that  their nth  derivative exists, then the nth derivative of  their product will  be

derivative of product of two functions by Leibnitz theorem-

Example-1 If   y  , then prove that-

Sol. Here it is given that-

On differentiating-

Or

= ny.2x

Differentiate again with respect to x, we get-

Or

   …………………. (1)

Differentiate each term of (1) by using Leibnitz’s theorem, we get-

Therefore we get-

Hence proved.

Example-2: If , then prove that-

Sol. Here we have-

Or

Or

y = b cos[ n log(x/n)]

On differentiating, we get-

Which becomes-

Differentiate again both sides with respect to x, we get-

It becomes-

       ……………….. (1)

Differentiate each term n times with respect to x, we get-

Which is-

Hence proved,

Example-3: If y = , then prove that-

Sol. It is given that-    y =

First derivative –

It becomes-

= 

=

=

Becomes-

+ - = 0

Om differentiating again we get-

+ = 0

Or

+ = 0

Differentiate n times by using Leibnitz’s theorem, we get-

So that

Hence proved.

Example-4: Find  the nth derivative of  

Sol.

Let

Also

By Leibnitz’s theorem

    …(i)

Here 

Differentiating with  respect  to  x, we get

Again differentiating  with respect to x, we get

Similarly the  nth derivative will be

From  (i) and (ii) we have,

Example-5: If ,     then show that

Sol.

Also, find 

Here

Differentiating with respect to x, we get

   …(ii)

Squaring   both  side  we get

   …(iii)

Again differentiating with  respect to  x ,we get

Using Leibnitz’s  theorem  we  get

        …(iv)

Putting x=0  in equation (i),(ii) and (iii)  we get

Putting  n=1,2,3,4….

………………

Hence   

Example-6: If then show that 

Sol.

Given 

Differentiating both side  with  respect to  x.

          …..(ii)

Again differentiating with  respect to  x, we get

     …(iii)

By  Leibnitz’s theorem

…(iv)

Putting x=0 in equation  (i),(ii),(iii) and (iv) we  get

Putting n=1,2,3,4… so  we get

Hence we  have

Key takeaways-

1. 
2. The differential coefficient is-
3. 1.
4. 2.
5. 3.
6. 4.
7. 5.
8. 6.
9. 7.

1.3 Expansion of a function into Taylors and Maclaurin’sSeries

Maclaurin’s series-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)  

We get-

+ …….

Expansion of functions using standard expansions-

Example: By using Maclaurin’s series expand tan x.

Sol.

Let-

Put these values in Maclaurin’s series we get-

Example: Expand by using Maclaurin’s series.

Sol.

Let

Put these values in Maclaurin’s series-

Or

Expand by Maclaurin’s theorem,

Log sec x

Solution:

Let f(x) = log sec x

By Maclaurin’s Expansion’s,

   (1)

By equation (1)

Prove that

Solution:

Here f(x)  = x cosec x

=

Now we know that

Q. Expand upto x6

Solution:

Here

Now we know that

    … (1)

    … (2)

Adding (1) and (2) we get

Q. Show that

Solution:

Here

Thus

Taylor’s Series Expansion:

a)     The expansion of f(x+h) in ascending power of x is

b)    The expansion of f(x+h) in ascending power of h is

c)     The expansion of f(x) in ascending powers of (x-a) is,

Using the above series expansion we get series expansion of f(x+h) or f(x).

Expansion of functions using standard expansions-

Example-1: Expand in power of (x – 3)

Solution:

Let

Here a = 3

Now by Taylor’s series expansion,

 … (1)

equation (1) becomes.

Example-2:

Using Taylors series method expand in powers of (x + 2)

Solution:

Here

a = -2

By Taylors series,

   … (1)

Since

,  , …..

Thus equation (1) becomes

Example-3:

Expand in ascending powers of x.

Solution:

Here

i.e.

Here h = -2

By Taylors series,

    … (1)

equation (1) becomes,

Thus

Example-4:

Expand in powers of x using Taylor’s theorem,

Solution:

Here

i.e.

Here

h = 2

By Taylors series

  … (1)

By equation (1)

Key takeaways-

Taylor’s Series Expansion:-

Taylor’s Series Expansion:-

a)     The expansion of f(x+h) in ascending power of x is

b)    The expansion of f(x+h) in ascending power of h is

c)     The expansion of f(x) in ascending powers of (x-a) is,

Using the above series expansion we get series expansion of f(x+h) or f(x).

Expansion of functions using standard expansions-

Example-1: Expand in power of (x – 3)

Solution:

Let

Here a = 3

Now by Taylor’s series expansion,

 … (1)

equation (1) becomes.

Example-2:

Using Taylors series method expand in powers of (x + 2)

Solution:

Here

a = -2

By Taylors series,

   … (1)

Since

,  , …..

Thus equation (1) becomes

Example-3:

Expand in ascending powers of x.

Solution:

Here

i.e.

Here h = -2

By Taylors series,

    … (1)

equation (1) becomes,

Thus

Example-4:

Expand in powers of x using Taylor’s theorem,

Solution:

Here

i.e.

Here

h = 2

By Taylors series

  … (1)

By equation (1)

Key takeaways-

Taylor’s Series Expansion:-

Exercise

a)     Expand in powers of (x – 2)

b)    Expand in powers of (x + 2)

c)     Expand in powers of (x – 1)

d)    Using Taylors series, express in ascending powers of x.

e)     Expand in powers of x, using Taylor’s theorem.

1.4 Function of two or more variables: Partial derivatives

Function of two variables- 

R denotes the set of real numbers. Let’s D is a collection of pairs of real numbers (x , y), which means D is a subset of R × R.

Then a real valued function of two variables of f is a rule that assign to each point (x,y) in D a unique real number denoted by f(x,y).

The set D is called the domain of f.

The set [ f(x,y) : (x,y) belongs to D ], which is the set of values the function f takes, is called range of f.

Here we genrally use the letter z to denote the value that a function of two variables takes,

Then we will have,

z = f(x, y)

Here we will call that z is the dependent variable and x and y are independent variables.

Example-1: The area of a rectangular figure whose length is l and breadth is b, given by  l × b.

Here independent variables are l and b, but dependent variable is area.

Example-2: the volume of a cylinder is given by πr²h, where r is the radius and h is the height of the cylinder.

In which radius and height are independent variables and volume is dependent.

Example-3:  the volume of cuboid is given by l × b× h. Where l, b and h are the length, breadth and height respectively.

L , b and h are independent variable and volume of cuboid is dependent variable.

Limits-

The function f(x,y) is said to tend to limit ‘l’ , as x →a and  y→b Iff the limit is dependent on point (x,y) as  x →a and  y→b

We can write this as,

Example-1: Evaluate the

Sol. We can simply find the solution as follows,

Example-2:  Evaluate

Sol.

-6.

Example-3: evaluate

Sol.

Conitinuity –

At point (a,b) , a function f(x,y) is said to be continuous if,

Working rule for continuity-

Step-1: f(a,b) should be well defined.

Step-2: should exist.

Step-3:

Example-1: Test the continuity of the following function-

Sol. (1) the function is well defined at (0,0)

(2) check for the second step,

That means the limit exists at (0,0)

Now check step-3:

So that the function is continuous at origin.

Example-2: check for the continuity of the following function at origin,

Sol. (1) Here the function is well defined at (0,0)

(2) check for second step-

Limit f is not unique for different values of m

So that the limit does not exists.

Therefore the function is not continuous at origin.

Steps to check for existence of limit-

Step-1: find the value of f(x,y) along x →a and  y→b

Step-2: find the value of f(x,y) along x →b and  y→a

Note- if the values in step -1 and step-2 are same then we can say that the limits exist otherwise not.

Step-3: if a →0 and  b→0 then find the limit along y =mx , if the value does not contain m then limit exist,  If it contains m then the limit does not exist.

Note-1- put x = 0 and y = 0 in f , then find f1

2 - Put y = 0 and x = 0 In f  then find x2

If f1 and f2 are equal then limit exist otherwise not.

3- put y = mx  then find f3

If f1 = f2 ≠f3, limit does not exist.

4- put y = mx² and find f4,

If f1 = f2 = f3 ≠ f4 , limit does not exist

If f1 = f2 = f3 = f4 , limit exist.

Example-1:  Evaluate 

Sol . 1.     

2.

Here f1 = f2

3. Now put y = mx, we get

Here f1 = f2 = f3

Now put y = mx²

4.

Therefore ,

F1 = f2 = f3 =f4

We can say that the limit exists with 0.

Example-2: evaluate the following-

Sol.  First we will calculate f1 –

Here we see that f1 = 0

Now find f2,

Here , f1 = f2

Therefore the limit exists with value 0.

Partial derivatives

First order 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.

Higher order partial differentiation-

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.

Example-1: Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

Sol. 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.

Example-2: Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

Sol. 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)

Example-3: Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy⁵)

Sol. 3x²y² - y⁵,      2x³y – 5xy⁴,

  = = 6xy²

  = 2x³ - 20xy³

= = 6x²y – 5y⁴

= = 6x²y -   5y⁴

Example-4: Find

Sol.  First we will differentiate partially with respect to r,

Now differentiate partially with respect to θ, we get

Example-5:  if,

Then find.

Sol-

We have

Example-6: if , then show that-

Sol. Here we have,

u =     …………………..(1)

Now partially differentiate eq.(1) w.r to x and y , we get

=

Or

                 ………………..(2)

And now,

=

         ………………….(3)

Adding eq. (1) and (3) , we get

  =  0

Hence proved.

Derivatives of composite and implicit functions

A composite function is a composition / combination of the functions. In this value of one function depends on the value of another function. A composite function is created when one function is put in another.

Let

i.e

To differentiate composite function chain rule is used:

Chain rule:

1. If where x,y,z are all the function of t then

2.     If be an implicit relation between x and y .

Differentiating with respect to x we get

We get           

Example1: If where then find the value of ?

Sol.

Given  

Where 

By chain rule

Now substituting the value of x ,y,z we get

-6

8

Example2: If then calculate

Sol

Given

By Chain Rule

Putting the value of u =

Again partially differentiating z with respect to y

By Chain Rule

by substituting value

Example 3 :If .

Show that   

Given

Partially differentiating u with respect to x and using chain rule

………(i)

Partially differentiating z with respect to y and using chain rule

=   ………..(ii)

Partially differentiating z with respect to t and using chain rule

Using (i) and (ii) we get

Hence   

Example4: If where the relation is .

Find the value of

Let the given relation is denoted by

We know that

Differentiating u with respect to x and using chain rule

Example5 : If and the relation is . Find

Given relation can be rewrite as

.

We know that

Differentiating u with respect to x and using chain rule

Implicit differentiation-

Let f(x,y) = 0

Where y = ∅(x)

By the chain rule , with x = x and y = ∅(x), we get

Here we assume that y is a differentiable funtion of x.

Example-1: if ∅ is a differentiable function such that y = ∅(x) satisfies the equation

x³ + y³ +sin xy = 0 then find .

Sol.  Suppose      f(x,y) =  x³ + y³ +sin xy

Then,

fᵡ =  3x² + y cos xy

Fy = 2y +  x cos xy

So,

Example-2:

Sol.  Take partial derivative on both side w.r. t. x , treat y as constant

Example-3:  if x²y³ + cos y cos z  =  x² cos x sin y, then find 

Sol. Differentiate partially w.r.t. x and treat y as constant,

Key takeaways-

1. The function f(x,y) is said to tend to limit ‘l’ , as x →a and  y→b Iff the limit is dependent on point (x,y) as  x →a and  y→b
2. 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:

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

(a)     =

(b)  =

(c)   =

(d)     =

1.5 Euler’s Theorem

Homogeneous 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.

Or

A function f(x,y) is said to be a homogenous function in which the power of each term is the same.

Example:

1. The function-

Is a homogeneous function of order 3.

Euler’s theorem

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,

= nf(y/x) + xⁿ f’(y/x).()

Now multiplying by x on both sides, we get

x = nf(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 = nf(y/x) + + xⁿ f’(y/x).()  + xⁿ f’(y/x).

xy         =  nf(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.

Let’s 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 – xysin(xy) + cos(xy)      -----------------(2)

From equation (1) and (2),

  = 

Example-3: If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,

Sol. Here we have,

u(x,y,z) = log( tan x + tan y + tan z)  ………………..(1)

Diff. Eq.(1) w.r.t. x , partially , we get

   ……………..(2)

Diff. Eq.(1) w.r.t. y , partially , we get

 ………………(3)

Diff. Eq.(1) w.r.t. z , partially , we get

     ……………………(4)

Now multiply eq. 2 , 3 , 4 by sin 2x , sin 2y , sin 2z respectively and adding , in order to get the final result,

We get,

=

So that, 

hence proved.

Key takeaways-

1. A function f(x,y) is said to be a homogenous function in which the power of each term is the same.
2. Euler’s theorem- if u = f(x, y) be a homogeneous function in x and y of degree n , then

x + y = nu

1.6 Taylor’s Expansion

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + ……..     (1)

Example-1: Express the polynomial in powers of (x-2).

Sol.  Here we have,

f(x) =

Differentiating the function w.r.t. x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

Now using Taylor’s theorem-

+ …….     (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) = 

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12   and f’’’’(2) = 0

Now put a = 2 and substitute the above values in equation(1), we get-

Taylor’s theorem for functions of two variables-

Suppose f(x , y) be a function of two independent variables x and y. Then,

+ ……………

Maclaurin’s series is the special case of Taylor’s series-

When we put a = 0 and b = 0 (about origin) in Taylor’s series, we get-

+ ……………

Example-2:  Expand f(x , y) = in powers of x and y about origin.

Sol. Here we have the function-

f(x , y) =

Here , a = 0 and b = 0 then

f(0 , 0) =

Now we will find partial derivatives of the function-

Now using Taylor’s theorem-

+………

Suppose h = x and k = y, we get

+…….

= +……….

Example-3: Find the Taylor’s expansion of about (1 , 1) up to second degree term.

Sol. We have,

At (1, 1)

Now by using Taylor’s theorem-

……

Suppose 1 + h = x  then h =  x – 1

1 + k = y   then k = y - 1

……

  =

……..

Key takeaways-

1. Taylor’s Theorem-

+ …….+ + ……..

2.     Maclaurin’s Theorem-

+ …….

1.7 Maxima & Minima of a function of two variables

Maxima and minima of function 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 complement 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 (

Key takeaways-

1. 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.
2. 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

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

At the point.

1.8 Lagrange’s method of undetermined multipliers

Let be a function of x, y, z which to be discussed for stationary value.

Let be a relation in x, y, z

for stationary values we have,

i.e.   … (1)

Also from we have

    … (2)

Let ‘’ be undetermined multiplier then multiplying equation (2) by and adding in equation (1) we get,

     … (3)

     … (4)

      … (5)

Solving equation (3), (4) (5) & we get values of x, y, z and .

Q1) Decampere a positive number ‘a’ in to three parts, so their product is maximum

S1)

Let x, y, z be the three parts of ‘a’ then we get.

    … (1)

Here we have to maximize the product

i.e.

By Lagrange’s undetermined multiplier, we get,

       … (2)

       … (3)

        … (4)

i.e.

        … (2)’

        … (3)’

        … (4)

And

From (1)

Thus .

Hence their maximum product is 

Q2) Find the point on plane nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.

S2)

Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition    … (1)

By method of Lagrange’s undetermined multipliers we have

       … (2)

       … (3)

i.e.&

       … (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

If where x + y + z = 1.

Prove that the stationary value of u is given by,

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 tozero

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 tozero

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

References:

1. G.B. Thomas and R.L. Finney, “Calculus and Analytic Geometry”, Pearson, 2002.

2. T. Veerarajan, “Engineering Mathematics”, McGraw-Hill, New Delhi, 2008.

3. B. V. Ramana, “Higher Engineering Mathematics”, McGraw Hill, New Delhi, 2010.

4. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

5. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

6. HK Dass