Partial differentiation

Chain Rule And Euler’s Theoram

G.H.Patel College of Engineering And Technology

Created By Enrolment number130110120022 To 130110120028

Tanuj Parikh Akash Pansuriya

The Chain Rule

In this section, we will learn

about:

The Chain Rule

thE CHAIN RULE

• Recall that the Chain Rule for functions of a single variable gives the following rule for differentiating a composite function.

• Suppose that z = f(x, y) is a differentiable function of x and y, where x = g(t) and y = h(t) are both differentiable functions of t.

– Then, z is a differentiable function of tand

THE CHAIN RULE (CASE 1)

dz f dx f dy

dt x dt y dt

• If y = f(x) and x = g(t), where f and g are differentiable functions, then y is indirectly a differentiable function of t, and

THE CHAIN RULE

dy dy dx

dt dx dt

• For functions of more than one variable, the Chain Rule has several versions.

– Each gives a rule for differentiating a composite function.

THE CHAIN RULE

-Example(1)

• The pressure P (in kilopascals), volume V(in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equation

PV = 8.31T


• Find the rate at which the pressure is changing when:

– The temperature is 300 K and increasing at a rate of 0.1 K/s.

– The volume is 100 L and increasing at a rate of 0.2 L/s.


• If t represents the time elapsed in seconds, then, at the given instant, we have:

– T = 300

– dT/dt = 0.1

– V = 100

– dV/dt = 0.2


• Since , the Chain Rule gives:

– The pressure is decreasing at about 0.042 kPa/s.


8.31T

PV

2

2

8.31 8.13

8.31 8.31(300)(0.1) (0.2)

100 100

0.04155

dP P dT P dV dT T dV

dt T dt V dt V dt V dt

• We now consider the situation where z = f(x, y), but each of x and y is a function of two variables s and t: x = g(s, t), y = h(s, t).

– Then, z is indirectly a function of s and t,and we wish to find ∂z/∂s and ∂z/∂t.


• Recall that, in computing ∂z/∂t, we hold s fixed and compute the ordinary derivative of z with respect to t.

– So, we can apply Theorem 2 to obtain:


z z x z y

t x t y t

• Suppose z = f(x, y) is a differentiable function of x and y, where x = g(s, t) and y = h(s, t) are differentiable functions of s and t.

– Then,


z z x z y z z x z y

s x s y s t x t y t

-Example(2)

• If z = ex sin y, where x = st2 and y = s2t, find ∂z/∂s and ∂z/∂t.

– Applying Case 2 of the Chain Rule, we get the following results.



2 2

2

2 2 2

( sin )( ) ( cos )(2 )

sin( ) 2 cos( )

x x

st st

z z x z y

s x s y s

e y t e y st

t e s t ste s t


2 2

2

2 2 2

( sin )(2 ) ( cos )( )

2 sin( ) cos( )

x x

st st

z z x z y

t x t y t

e y st e y s

ste s t s e s t

• Case 2 of the Chain Rule contains three types of variables:

– s and t are independent variables.

– x and y are called intermediate variables.

– z is the dependent variable.

THE CHAIN RULE

• To remember the Chain Rule, it’s helpful to draw a tree diagram, as follows.

THE CHAIN RULE

• We draw branches from the dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y.

TREE DIAGRAM

• Then, we draw branches from x and yto the independent variables s and t.

– On each branch, we write the corresponding partial derivative.

TREE DIAGRAM

• To find ∂z/∂s, we find the product of the partial derivatives along each path from z to s and then add these products:

TREE DIAGRAM

z z x z y

s x s y s

• Similarly, we find ∂z/∂t by using the paths from z to t.

TREE DIAGRAM

• Suppose u is a differentiable function of the n variables x1, x2, …, xn and each xj

is a differentiable function of the m variables t1, t2 . . . , tm.

THE CHAIN RULE (GEN. VERSION)

• Then, u is a function of t1, t2, . . . , tm

and

• for each i = 1, 2, . . . , m.


1 2

1 2

n

i i i n i

xx xu u u u

t x t x t x t

-Example(3)

• If u = x4y + y2z3, where x = rset, y = rs2e–t, z = r2s sin t

find the value of ∂u/∂s when r = 2, s = 1, t = 0


• With the help of this tree diagram, we have:


3 4 3 2 2 2(4 )( ) ( 2 )(2 ) (3 )( sin )t t

u

s

u x u y u z

x s y s z s

x y re x yz rse y z r t

• When r = 2, s = 1, and t = 0, we have:

x = 2, y = 2, z = 0

• Thus,


(64)(2) (16)(4) (0)(0)

192

u

s

29

Homogeneous Functions

• A function f(x1,x2,…xn) is said to be homogeneous of degree k if

f(tx1,tx2,…txn) = tk f(x1,x2,…xn)

– when a function is homogeneous of degree one, a doubling of all of its arguments doubles the value of the function itself

– when a function is homogeneous of degree zero, a doubling of all of its arguments leaves the value of the function unchanged

30

Homogeneous Functions

• If a function is homogeneous of degree k, the partial derivatives of the function will be homogeneous of degree k-1

31

Euler’s Theorem

• If we differentiate the definition for homogeneity with respect to the proportionality factor t, we get

ktk-1f(x1,…,xn) = x1f1(tx1,…,txn) + … + xnfn(x1,…,xn)

• This relationship is called Euler’s theorem

32

Euler’s Theorem

• Euler’s theorem shows that, for homogeneous functions, there is a definite relationship between the values of the function and the values of its partial derivatives

Fall 2002

This definition can be further enlarged to include transcendental functions also as follows. A function f(x,y) is said to be homogeneous function of degree n if it can be expressed as

OR

x

yxn

y

xyn

Fall 2002

Theorem: if f is a homogeneous function of x and y of degree n then

Cor: if f is a homogeneous function of degree n, then

nfy

fy

x

fx

fnny

fy

yx

fxy

x

fx )1(2

2

22

2

2

22

Fall 2002

Euler’s theorem for three variables: If f is a homogeneous function of three independent variables x, y, z of order n, then

nfzfyfxf zyx

Modified EULER’s theorem

)('

)(

uf

ufn

y

uy

x

ux

If Z is a Homogeneous function of degree n in

the variables x and y and z=f(u)

If z is a homogeneous function of degree n in

the variables x and y and z=f(u)

],1)(')[(22

2

22

2

22

ugug

yx

uxy

y

uy

x

ux

)('

)()(

uf

ufnug

where