partial derivatives 15. 2 15.5 the chain rule in this section, we will learn about: the chain rule...

PARTIAL DERIVATIVESPARTIAL DERIVATIVES

15

2

15.5The Chain Rule

In this section, we will learn about:

The Chain Rule and its application

in implicit differentiation.

PARTIAL DERIVATIVES

3

Recall that the Chain Rule for functions

of a single variable gives the following rule

for differentiating a composite function.

THE CHAIN RULE

4

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

Equation 1

5

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

6

The first version (Theorem 2) deals with

the case where z = f(x, y) and each of

the variables x and y is, in turn, a function

of a variable t.

This means that z is indirectly a function of t, z = f(g(t), h(t)), and the Chain Rule gives a formula for differentiating z as a function of t.

THE CHAIN RULE

7

We assume that f is differentiable

(Definition 7 in Section 15.4).

Recall that this is the case when fx and fy are continuous (Theorem 8 in Section 14.4).

THE CHAIN RULE

8

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 t and

THE CHAIN RULE (CASE 1) Theorem 2

dz f dx f dy

dt x dt y dt

9

A change of ∆t in t produces changes

of ∆x in x and ∆y in y.

These, in turn, produce a change of ∆z in z.

THE CHAIN RULE (CASE 1) Proof

10

Then, from Definition 7 in Section 15.4,

we have:

where ε1 → 0 and ε2 → 0 as (∆x, ∆y) → (0,0).

If the functions ε1 and ε2 are not defined at (0, 0), we can define them to be 0 there.


1 2

f fz x y x y

y y

11

Dividing both sides of this equation by ∆t,

we have:

THE CHAIN RULE (CASE 1)

z

t

f

x

x

t

f

y

y

t

1

x

t

2

y

t

Proof

12

If we now let ∆t → 0 , then

∆x = g(t + ∆t) – g(t) → 0

as g is differentiable and thus continuous.

Similarly, ∆y → 0.

This, in turn, means that ε1→ 0 and ε2→ 0.


13

Thus,


0

10 0 0 0

20 0

lim

lim lim lim lim

lim lim

0 0

t

t t t t

t t

dz z

dt tf x f y x

x t y t t

y

tf dx f dy dx dy

x dt y dt dt dt

f dx f dy

x dt y dt

Proof

14

Since we often write ∂z/∂x in place of ∂f/∂x,

we can rewrite the Chain Rule in the form


dz z dx z dy

dt x dt y dt

15

If z = x2y + 3xy4, where x = sin 2t and

y = cos t, find dz/dt when t = 0.

The Chain Rule gives:


4 2 3(2 3 )(2cos 2 ) ( 12 )( sin )

dz z dx z dy

dt x dt y dt

xy y t x xy t

Example 1

16

It’s not necessary to substitute the

expressions for x and y in terms of t.

We simply observe that, when t = 0, we have x = sin 0 = 0 and y = cos 0 = 1.

Thus,

THE CHAIN RULE (CASE 1) Example 1

0

(0 3)(2cos 0) (0 0)( sin 0) 6t

dz

dt

17

The derivative in Example 1 can be

interpreted as:

The rate of change of z with respect to t as the point (x, y) moves along the curve C with parametric equations x = sin 2t, y = cos t


Fig. 15.5.1, p. 938

18

In particular, when t = 0,

The point (x, y) is (0, 1).

dz/dt = 6 is the rate of increase as we move along the curve C through (0, 1).


Fig. 15.5.1, p. 938

19

If, for instance, z = T(x, y) = x2y + 3xy4

represents the temperature at the point (x, y),

then

The composite function z = T(sin 2t, cos t) represents the temperature at points on C

The derivative dz/dt represents the rate at which the temperature changes along C.


20

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


21

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.


22

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


23

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 1000.04155

dP P dT P dV dT T dV

dt T dt V dt V dt V dt

Example 2

24

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.


25

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

26

A similar argument holds for ∂z/∂s.

So, we have proved the following

version of the Chain Rule.


27

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,

THE CHAIN RULE (CASE 2) Theorem 3

z z x z y z z x z y

s x s y s t x t y t

28

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.


29


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

30


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

31

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

32

Notice that Theorem 3 has one term

for each intermediate variable.

Each term resembles the one-dimensional Chain Rule in Equation 1.

THE CHAIN RULE

33

To remember the Chain Rule,

it’s helpful to draw a tree diagram,

as follows.

THE CHAIN RULE

34

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

Fig. 15.5.2, p. 939

35

Then, we draw branches from x and y

to the independent variables s and t.

On each branch, we write the corresponding partial derivative.

TREE DIAGRAM

Fig. 15.5.2, p. 939

36

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

Fig. 15.5.2, p. 939

37

Similarly, we find ∂z/∂t by using

the paths from z to t.

TREE DIAGRAM

Fig. 15.5.2, p. 939

38

Now, we consider the general situation in

which a dependent variable u is a function

of n intermediate variables x1, . . . , xn.

Each of this is, in turn, a function of m

independent variables t1 , . . ., tm.

THE CHAIN RULE

39

Notice that there are n terms—one for

each intermediate variable.

The proof is similar to that of Case 1.

THE CHAIN RULE

40

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.

Theorem 4THE CHAIN RULE (GEN. VERSION)

41

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

and

for each i = 1, 2, . . . , m.

THE CHAIN RULE (GEN. VERSION)

1 2

1 2

n

i i i n i

xx xu u u u

t x t x t x t

Theorem 4

42

Write out the Chain Rule for the case

where w = f(x, y, z, t)

and

x = x(u, v), y = y(u, v), z = z(u, v), t = t(u, v)

We apply Theorem 4 with n = 4 and m = 2.

THE CHAIN RULE (GEN. VERSION) Example 4

43

The figure shows the tree diagram.

We haven’t written the derivatives on the branches.

However, it’s understood that, if a branch leads from y to u, the partial derivative for that branch is ∂y/∂u.


Fig. 15.5.3, p. 940

44

With the aid of the tree diagram, we can now

write the required expressions:


w w x w y w z w t

u x u y u z u t u

w w x w y w z w t

v x v y v z v t v

Example 4

45

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


46

With the help of

this tree diagram,

we have:


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

u

su x u y u z

x s y s z s

x y re x yz rse y z r t

Example 5

Fig. 15.5.4, p. 940

47

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

Example 5

48

If g(s, t) = f(s2 – t2, t2 – s2) and f is

differentiable, show that g satisfies

the equation


0g g

t ss t

49

Let x = s2 – t2 and y = t2 – s2.

Then, g(s, t) = f(x, y), and the Chain Rule gives:


(2 ) ( 2 )

( 2 ) (2 )

g f x f y f fs s

s x s y s x y

g f x f y f ft t

t x t y t x y

Example 6

50

Therefore,


2 2 2 2

0

g gt s

s t

f f f fst st st st

x y x y

Example 6

51

If z = f(x, y) has continuous second-order

partial derivatives and x = r2 + s2 and y = 2rs,

find:

a. ∂z/∂r

b. ∂2z/∂r2


52

The Chain Rule gives:

THE CHAIN RULE (GEN. VERSION) Example 7 a

(2 ) (2 )

z z x z y

r x r y r

z zr s

x y

53

Applying the Product Rule to the expression

in part a, we get:

THE CHAIN RULE (GEN. VERSION) E. g. 7 b—Equation 5

2

22 2

2 2 2

z z zr s

r r x y

z z zr s

x r x r y

54

However, using the Chain Rule again,

we have the following results.

THE CHAIN RULE (GEN. VERSION) Example 7 b

Fig. 15.5.5, p. 941

55


2 2

2

2 2

2

(2 ) (2 )

(2 ) (2 )

z z x z y

r x x x r y x r

z zr s

x y x

z z x z y

r y x y r y y r

z zr s

x y y

Example 7 b

56

Putting these expressions into Equation 5 and

using the equality of the mixed second-order

derivatives, we obtain the following result.

THE CHAIN RULE (GEN. VERSION) Example 7 b

57


2 2 2

2 2

2 2

2

2 2 22 2

2 2

2 2 2 2

2 2 2

2 4 8 4

z z z zr r s

r x x y x

z zs r s

x y y

z z z zr rs s

x x x y y

Example 7 b

58

The Chain Rule can be used to give

a more complete description of the process

of implicit differentiation that was introduced

in Sections 3.6 and 15.3

IMPLICIT DIFFERENTIATION

59

We suppose that an equation of

the form F(x, y) = 0 defines y implicitly

as a differentiable function of x.

That is, y = f(x), where F(x, f(x)) = 0 for all x in the domain of f.


60

If F is differentiable, we can apply Case 1 of

the Chain Rule to differentiate both sides of

the equation F(x, y) = 0 with respect to x.

Since both x and y are functions of x, we obtain:


0F dx F dy

x dx y dx

61

However, dx/dx = 1.

So, if ∂F/∂y ≠ 0, we solve for dy/dx

and obtain:


x

y

FFdy x

Fdx Fy

Equation 6

62

To get the equation, we assumed F(x, y) = 0

defines y implicitly as a function of x.

The Implicit Function Theorem, proved

in advanced calculus, gives conditions under

which this assumption is valid.

IMPLICIT FUNCTION THEOREM

63

The theorem states the following.

Suppose F is defined on a disk containing (a, b), where F(a, b) = 0, Fy(a, b) ≠ 0, and Fx and Fy are continuous on the disk.

Then, the equation F(x, y) = 0 defines y as a function of x near the point (a, b) and the derivative of this function is given by Equation 6.


64

Find y’ if x3 + y3 = 6xy.

The given equation can be written as:

F(x, y) = x3 + y3 – 6xy = 0

So, Equation 6 gives:

IMPLICIT DIFFERENTIATION Example 8

2 2

2 2

3 6 2

3 6 2x

y

Fdy x y x y

dx F y x y x

65

Now, we suppose that z is given implicitly

as a function z = f(x, y) by an equation of

the form F(x, y, z) = 0.

This means that F(x, y, f(x, y)) = 0 for all (x, y) in the domain of f.


66

If F and f are differentiable, then we can use

the Chain Rule to differentiate the equation

F(x, y, z) = 0 as follows:


0F x F y F z

x x y x z x

67

However,

So, that equation becomes:


( ) 1 and ( ) 0x yx x

0F F z

x z x

68

If ∂F/∂z ≠ 0, we solve for ∂z/∂x and obtain

the first formula in these equations.

The formula for ∂z/∂y is obtained in a similar manner.


FFz z yx

F Fx yz z

Equations 7

69

Again, a version of the Implicit Function

Theorem gives conditions under which

our assumption is valid.


70

This version states the following.

Suppose F is defined within a sphere containing (a, b, c), where F(a, b, c) = 0, Fz(a, b, c) ≠ 0, and Fx, Fy, and Fz are continuous inside the sphere.

Then, the equation F(x, y, z) = 0 defines z as a function of x and y near the point (a, b, c), and this function is differentiable, with partial derivatives given by Equations 7.


71

Find ∂z/∂x and ∂z/∂y if

x3 + y3 + z3 + 6xyz = 1

Let F(x, y, z) = x3 + y3 + z3 + 6xyz – 1

IMPLICIT DIFFERENTIATION Example 9

72

Then, from Equations 7, we have:


2 2

2 2

2 2

2 2

3 6 2

3 6 2

3 6 2

3 6 2

x

z

y

z

Fz x yz x yz

x F z xy z xy

Fz y xz y xz

y F z xy z xy

Example 9

partial derivatives 15. 2 15.5 the chain rule in this section, we will learn about: the chain rule...

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