Unit-4

Multivariable Differential Calculus

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)

= 2xycos(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 repsect to r,

Now differentiate partially with respect to θ, we get

Example-5: if,

Then find.

Sol-

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

Hence proved.

Key takeaways-

1.

(a) =

(b) =

(c) =

(d) =

When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant.

Let the function, u = f( x, y), such that x = g(t) , y = h(t)

ᵡ Then we can write,

=

=

This is the total derivative of u with respect to t.

Change of variable-

If w = f (x, y) has continuous partial variables fxand fyand if x = x (t), y = y (t) are

Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a

Differentiable function of t.

In this case, we get,

fx(x (t), y (t)) x’(t)+ fy(x(t), y (t)) y’(t).

Example-:1 let q = 4x + 3y and x = t³ + t² + 1 , y = t³ - t² - t

Then find .

Sol. :. =

Where, f1 = , f2 =

In this example f1 = 4 , f2 = 3

Also, 3t² + 2t ,

4(3t² + 2t) + 3(

= 21t² + 2t – 3

Example-2: Find if u = x³y⁴ where x = t³ and y = t².

Sol. As we know that by definition, =

3x²y⁴3t² + 4x³y³2t = 17t¹⁶.

Example-3: if w = x² + y – z + sintand x + y = t, find

(a) y,z

(b) t, z

Sol. With x, y, z independent, we have

t = x + y, w = x²+ y - z + sin (x + y).

Therefore,

y,z = 2x + cos(x+y)(x+y)

= 2x + cos (x + y)

With x, t, z independent, we have

Y = t-x, w= x² + (t-x) + sin t

Thust, z = 2x - 1

Example-4: If u = u( y – z , z - x , x – y) then prove that = 0

Sol. Let,

Then,

By adding all these equations we get,

= 0 hence proved.

Example-5: if φ( cx – az , cy – bz) = 0 then show that ap + bq = c

Where p = q =

Sol. We have,

φ( cx – az , cy – bz) = 0

φ( r , s) = 0

Where,

We know that,

Again we do,

By adding the two results, we get

Example-6: If z is the function of x and y , and x = , y = , then prove that,

Sol. Here , it is given that, z is the function of x and y & x , y are the functions of u and v.

So that,

……………….(1)

And,

………………..(2)

Also there is,

x = and y = ,

Now,

, , ,

From equation(1) , we get

……………….(3)

And from eq. (2) , we get

…………..(4)

Subtracting eq. (4) from (3), we get

= ) – (

= x

Hence proved.

Key takeaways:

- Let the function, u = f( x, y), such that x = g(t) , y = h(t)

ᵡ Then we can write,

=

=

This is the total derivative of u with respect to t.

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:

- 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 ?

Given

Where

By chain rule

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

-6

8

Example2 :If then calculate

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,

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.

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.

Example: Show that

Given

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

Example: 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

Example: If , find the value of

Given

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

Therefore by Euler’s theorem

Example: If

Given

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

Therefore by Euler’s theorem

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

Example: 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-

- A function f(x,y) is said to be a homogenous function in which the power of each term is the same.
- If u = f(x, y) be a homogeneous function in x and y of degree n , then

x + y = nu

References

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers

4. HK dass, engineering mathematics.