3.1 Partial derivatives Total derivative | unit 3 differential calculus ii

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Module 3

Differential Calculus-II

3.1 Partial derivatives, Total derivative

Partial Differentiation

Prove that

Partial differentiation of function of function

If z = f(u) and . Then z becomes a function of x & y. In this case, z becomes a function of x & y.

i.e.

Then

Similarly

Then z becomes a function of x, y & z.

…………….

where

Prove that

2. If V = show that

3. If show that

4. If then prove that

Partial Differentiation of composite function

a) Let and , then z becomes a function of , In this case, z is called a composite function of .

i.e.

b) Let possess continuous partial derivatives and let possess continuous partial derivatives, then z is called a composite function of x and y.

i.e.

Continuing in this way, …..

Ex. If Then prove that

Ex. If then prove that

Where is the function of x, y, z.

Ex. If where ,

then show that,

ii)

Notations of partial derivatives of the variable to be treated as a constant

Let

and

i.e.

Then means the partial derivative of u w.r.t. x treating y const.

To find from given reactions we first express x in terms of u & v.

i.e. & then diff. x w.r.t. u treating v constant.

To find express v as a function of y and u i.e. then diff. v w.r.t. y treating u as a const.

Ex. If , then find the value of

Ex. If , then prove that

3.2 Euler’s Theorem for homogeneous functions

Euler’s Theorem on Homogeneous functions:

Statement:

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

Deductions from Euler’s theorem

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

2. If be a homogeneous function 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

3.3 Taylor and Maclaurin’s theorems for a function of one and 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

example 2:

compute the Taylor series centerd at zero for f(x)= sinx

solution:

f(x)=sinx f(0)=0

f’(x)=cosx f’(0)=1

f’’(x)=-sinx f’’(0)=0

f’’’(x)= -cosx f’’’(0)=-1

f(4)(x)= cosx f(4) (0)= 1

applying taylor series we get

T(x) = = = x-

Thus turns out to converge x to sinx.

Maclaurian series:

Example 3:

(x)n

Example:

f(x)=

= f(0)+f’(0)x+ x2 + x3 +......

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

Example 4:

Find the maclurian series for f(x)= ex

Solution:

To get maclurian series,we look at the Taylor series polynomials for f near 0 and let them keep going.

Considering for example

By maclurian series we get,

3.4 Maxima and Minima of functions of several variables

Example 1:

Find out the maxima and minima of the function

Solution:

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

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

Example 2:

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 (

Example 3:

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

3.5 Lagrange Method of Multipliers, Jacobians

Example 1:

Determine the Jaccobian matrix of the function f: given by f(x,y,z)=(xy+2yz,2xy2z).

Solution:

We first f = () where given by f(x,y,z) = xy+2yz and = 2xwe now compute the gradients of the following:

Then the Jacobian matrix is given by,

Df(x,y,z) =

Example 2:

Determine the Jacobian matrix of the function defined for all

x= ( ) we have that,

f(x) = f(

Since f is real valued function the Jacobian f is simply the gradient of f.the gradient of f is given by

= Therefore, the Jacobian f defined whenever x

Example 3:

Suppose you are running a factor,producing some sort of widget that requires steel as a raw material.your costs are predominantly human labour,which is per hour for your workers,and the steel itself,which runs for $ 170 per ton.Suppose your revenue R is loosely modeled by the following equation: