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1/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiationMATH 209Calculus, III

Volker Runde

University of Alberta

Edmonton, Fall 2011


2/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Chain rule, first version

Theorem (chain rule, version I)

Let z=f(x, y) be a differentiable function of x and y, and letx=g(t) and y=h(t) be differentiable functions of t. Then zis a differentiable function of t with

dz

dt =

f

x

dx

dt +

f

y

dy

dt.


3/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, I

ExampleFind dz

dt

t=0

for

z=x2y+ 3xy4, x= sin(2t), y= cos t.

We have:

z

x = 2xy+ 3y4,

z

y

=x2 + 12xy3,

dx

dt = 2 cos(2t),

dy

dt = sin t.


4/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, II

Example (continued)

The chain rule yields:

dzdt

= zx

dxdt

+ zy

dydt

= (2xy+ 3y4)(2 cos(2t)) + (x2 + 12xy3)( sin t).

For t= 0: x= 0 and y= 1.

Hence:dz

dt

t=0

= 3 2 + 0 0 = 6.


5/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Chain rule, second version

Theorem (chain rule, version II)

Let z=f(x, y) be a differentiable function of x and y, and letx=g(s, t) and y=h(s, t) be differentiable functions of s andt. Then z is a differentiable function of s and t with

z

s =

z

x

x

s +

z

y

y

s

and z

t =

z

x

x

t +

z

y

y

t.


6/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, III

Example

Letz=ex sin y, x=st2, y=s2t.

Then

z

s =

z

x

x

s +

z

y

y

s

= (ex

sin y)t2

+ (ex

cos y)(2st)=t2est

2

sin(s2t) + 2stest2

cos(s2t)


7/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, IV

Example (continued)

. . .and

z

t =

z

x

x

t +

z

x

x

t

= (ex sin y)2st+ (ex cos y)s2

= 2stest2

sin(s2

t) +s2

e

st2

cos(s2

t).


8/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, V

Example

Letz=f(x, y), x=rcos , y=rsin .

Then

z

r =

z

x

x

r +

z

y

y

r =

z

xcos +

z

y sin

andz

=

z

x

x

+

z

y

y

=

z

xrsin +

z

yrcos .


9/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Chain rule, general version

Theorem (chain rule, version III)

Let u be a differentiable function of n variables x1, . . . , xn, and

let each xkbe a differentiable function of m variables

t1, . . . , tm. Then u is a differentible function of t1, . . . , tm such

thatu

tj=

u

x1

x1

tj+

u

x2

x2

tj+ +

u

xn

xn

tj

for j= 1, 2, . . . , m.


10/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, VI

ExampleLet

u=x4y+y2z3, x=rset, y=rs2et, z=rssin t.

Then

u

s =

u

x

x

s +

u

y

y

s +

u

z

z

s

= (4x3y)(ret) + (x4 + 2yz3)(2rset) + (3y2z2)(rsin t).

For r= 2, s= 1, and t= 0: x= 2, y= 2, and z= 0, so that

u

s

r=2,s=1,t=0

= 64 2 + 16 4 + 0 0 = 192.


11/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, VII

Example

Let z=f(x, y) have continuous second order derivatives, and

let x=r2 +s2 and y= 2rs. We want to find zr

and 2z

r2.

The chain rule yields

z

r =

z

x

x

r +

z

y

y

r =

z

x2r+

z

y2s

so that

2zr2

= r

2rz

x + 2sz

y

= 2z

x + 2r

r

z

x

+ 2s

r

z

y

.


12/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, VIII

Example (continued)

The chain rule again yields

r

z

x

=

x

z

x

x

r +

y

z

x

y

r

= 2z

x22r+

2z

yx2s

and

r

zz

=

x

zy

xr

+ y

zy

yr

= 2z

xy2r+

2z

y22s.


13/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, XI

Example (continued)

Eventually we obtain:

2z

r2 = 2

z

x + 2r

r z

x+ 2s

r z

y= 2

z

x + 2r

2r

2z

x2+ 2s

2z

yx

+ 2s2r

2z

xy + 2s

2z

y2= 2

z

x + 4r2

2z

x2+ 8rs

2z

xy + 4s2

2z

y2.


14/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Implicit functions, I

The two variable case

Let ybe a function ofx implicitlydefined by F(x, y) = 0.We obtain by the chain rule:

0 = Fx

dxdx

+ Fy

dydx

= F

x +

F

y

dy

dx.

If Fy = 0:dy

dx =

FxFy

.


15/17


16/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Implicit functions, II

The three variable caseLet z=f(x, y) be implicitly defined by F(x, y, z) = 0.We obtain by the chain rule:

0 = F

x

x

x=1

+F

y

y

x=0

+F

z

z

x

If Fz

= 0:

z

x =

Fx

Fz

.

Similarly (if Fz

= 0):

z

y =

Fy

Fz .


17/17

MATH 209

Calculus,III

Volker Runde

The chain rule

Implicit

differentiation

Examples, XI

Example

Find zx

and zy

ifx3 +y3 +z3 + 6xyz= 1.We have

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

so thatz

x =

3x2 + 6yz

3z2 + 6xy =

x2 + 2yz

z2 + 2xy

andz

y =

3y2 + 6xz

3z2 + 6xy =

y2 + 2xz

z2 + 2xy.

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