M38 Lec 021114

MATH 38Mathematical Analysis III

I. F. Evidente

IMSP (UPLB)

Outline

1 Implicit Differentiation

2 The Chain Rule

3 Differentiability

4 The Total Differential

Figures taken from: J. Stewart, The Calculus: Early Transcendentals,Brooks/Cole, 6th Edition, 2008.

Outline


2 The Chain Rule

3 Differentiability


Let z be a function in x and y defined implicitly by the equationF (x, y,z)=C , where C R.

ExampleWhat two functions are defined implicitly by x2+ y2+ z2 = 1?

z =1x2 y2

Two functions:

z =1x2 y2 and z =

1x2 y2

ExampleTwo functions defined implicitly by x2+ y2+ z2 = 1:

z =1x2 y2 and z =

1x2 y2

Partial derivatives with respect to x:For z =

1x2 y2:

z

x= x

1x2 y2

For z =1x2 y2:

z

x= x

1x2 y2


z =1x2 y2 and z =

1x2 y2

Partial derivatives with respect to x:

For z =1x2 y2:

z

x= x

1x2 y2

For z =1x2 y2:

z

x= x

1x2 y2


z =1x2 y2 and z =

1x2 y2

Partial derivatives with respect to x:For z =

1x2 y2:

z

x= x

1x2 y2

For z =1x2 y2:

z

x= x

1x2 y2

How can we get the partial derivatives of z without isolating z?

AnswerImplicit differentiation!

NoteSuppose z is a function of x and y . By the chain rule:

x[ f (z)]= d f

dz zx

ExampleSuppose z is a function of x and y .

1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz

zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]=

2y +4z zx

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +

4z zx

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z

zx

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]

= y3 3z2 zy

+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]=

y3 3z2 zy

+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3

3z2 zy

+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2

zy

+ z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+

z3 3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+ z3

3y2


x[ f (z)]= d f

dz zx


1

x[2xy +2z2]= 2y +4z z

x

2

y[y3z3]= y3 3z2 z

y+ z3 3y2

ExampleLet z be a function in x and y defined implicitly by x2+ y2+ z2 = 1. Findz

xand

z

y.

Note: If z =1x2 y2, then

z

x= x

z= x

1x2 y2

If z =1x2 y2, then

z

x= x

z= x

1x2 y2


xand

z

y.


z

x= x

z=

x1x2 y2

If z =1x2 y2, then

z

x= x

z= x

1x2 y2


xand

z

y.


z

x= x

z= x

1x2 y2

If z =1x2 y2, then

z

x= x

z= x

1x2 y2


xand

z

y.


z

x= x

z= x

1x2 y2

If z =1x2 y2, then

z

x= x

z=

x1x2 y2


xand

z

y.


z

x= x

z= x

1x2 y2

If z =1x2 y2, then

z

x= x

z= x

1x2 y2

Example

Let z be a function defined implicitly by ln(x+ z)= xy . Find zx

,z

yand

2z

xy.

Answers:

z

x= xy + yz1

z

y= x2+xz

Outline


2 The Chain Rule

3 Differentiability


Composition of FunctionsType 0: z = f (t ) and t = g (x, y)Type 1: z = f (x, y), x = x(t ), y = y(t )Type 2: z = f (x, y) and x = x(u,v), y = y(u,v)

Composition of Functions1 z(t )= ln t , t (x, y)= x2+ y2

z(x, y)= ln(x2+ y2)

z is a function of two variables x and y

2 z(x, y)= 3cosx sin(xy), x(t )= 1t, y(t )= 3t

z(t )= 3cos 1t sin(1

t3t)= 3cos 1

t sin3

z is a function of a single variable t3 z(x, y)= 8x2y 2x+3y , x(u,v)= uv , y(u,v)= u v

z(u,v)= 8u2v2(u v)2(u v)+3(u v)

z is a function of two variables u and v .


z(x, y)= ln(x2+ y2)



z(t )=

3cos1

t sin(1

t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)



z(t )= 3cos 1t

sin

(1

t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)=

3cos1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3

z is a function of a single variable t

3 z(x, y)= 8x2y 2x+3y , x(u,v)= uv , y(u,v)= u v

z(u,v)= 8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)=

8u2v2(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2

(u v)2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)

2(u v)+3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+

3(u v)



z(x, y)= ln(x2+ y2)




t3t)= 3cos 1

t sin3


z(u,v)= 8u2v2(u v)2(u v)+3(u v)


Recall:Type 0: f is a function of a single variable and g a function of twovariables x and y . Then

x

[f (g (x, y))

]= f (g (x, y)) x

[g (x, y)]

ExampleGiven f (x, y)= ln(x2+ y2), find fx(x, y).

Answer: fx(x, y)= 1x2+ y2

x(x2+ y2)= 1

x2+ y2 2x


x

[f (g (x, y))

]= f (g (x, y)) x

[g (x, y)]


Answer: fx(x, y)=

1

x2+ y2

x(x2+ y2)= 1

x2+ y2 2x


x

[f (g (x, y))

]= f (g (x, y)) x

[g (x, y)]



x(x2+ y2)= 1

x2+ y2 2x


x

[f (g (x, y))

]= f (g (x, y)) x

[g (x, y)]



x(x2+ y2)=

1

x2+ y2 2x


x

[f (g (x, y))

]= f (g (x, y)) x

[g (x, y)]



x(x2+ y2)= 1

x2+ y2 2x

Type 1: z = f (x, y), x = x(t ), y = y(t )TheoremIf x(t ) and y(t ) are differentiable at t and z = f (x, y) is differentiable at(x(t ), y(t )), then

dz

dt= zx

dxdt

+ zy

dydt

Note:For Type 1, note that z = f (x, y)= f (x(t ), y(t )) is a function of a singlevariable t . Thus, we call

dz

dtthe total derivative of z with respect to t

(as opposed to partial derivatives).

Example

z = 3cosx sin(xy), x = 1t, y = 3t

Type 2: z = f (x, y) and x = x(u,v), y = y(u,v)TheoremIf x = x(u,v) and y = y(u,v) have first order partial derivatives at (u,v) andif f (x, y) is differentiable at (x, (u,v), y(u,v)), then:

z

u= zx

xu

+ zy

yu

andz

v= zx

xv

+ zy

yv

Examplez = 8x2y 2x+3y , x = uv , y = u v

Extension to 3 or more variables

Exampleu = r s2 ln t , r = x2, s = 4y +1, t = xy3

Outline


2 The Chain Rule

3 Differentiability


Differentiability for y = f (x)f (x) is differentiable at x = x0 if f (x0) exists.

Properties of Differentiability1 The graph of f has a non-vertical tangent line at x = x02 The tangent line gives a local linear approximation at x0

f (x) f (x0)+ f (x0)(xx0)

when x is very close to x0.(Note that the local linear approximation is the first degree TaylorPolynomial at x = x0!)

3 f is continuous at x0

RecallThe graph of a differentiable function is "smooth".


Properties of Differentiability1 The graph of f has a non-vertical tangent line at x = x0

2 The tangent line gives a local linear approximation at x0

f (x) f (x0)+ f (x0)(xx0)






f (x) f (x0)+ f (x0)(xx0)






f (x) f (x0)+ f (x0)(xx0)

when x is very close to x0.

(Note that the local linear approximation is the first degree TaylorPolynomial at x = x0!)





f (x) f (x0)+ f (x0)(xx0)




We want to define differentiability for functions of two variables in such away that the a differentiable function also has those three properties.

Problem:The existence of fx(x0, y0) and fy (x0, y0) does not guarantee that f hasthose three properties!

DefinitionA function of 2 variables f (x, y) is said to be differentiable at (x0, y0) ifthe following conditions are satisfied:

1 fx(x0, y0) and fy (x0, y0) exist, and

2 lim(x,y)(x0,y0)

f (x, y)L(x, y)(xx0)2+ (y y0)2

= 0, where

L(x, y)= f (x0, y0)+ fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)

If f is differentiable at (x0, y0), L(x, y) is called the local linearapproximation of f at (x0, y0)

Examplef (x, y)= x2+ y2 is differentiable at (0,0).

DefinitionA function f (x, y) is said to be differentiable if f (x, y) is differentiable forall (x, y) R2.

TheoremIf a function is differentiable at a point, then it is continuous at that point.

RemarkProperties of a function f that is differentiable at (x0, y0):

1 The graph of f has a non-vertical tangent plane at (x0, y0, f (x0, y0)).An equation of this tangent plane is z = L(x, y).

2 L(x, y) approximates f (x, y) when (x, y) is very close to (x0, y0).

f (x, y) L(x, y)

3 f is continuous at (x0, y0).

TheoremIf all first-order partial derivatives of f exist and are continuous at a point,then f is differentiable at that point.

ExampleBy the theorem, f (x, y)= x2+ y2 is differentiable.

RemarkAlso by the theorem:

1 A polynomial functions are differentiable.2 A rational functions is differentiable at every point in its domain.

Outline


2 The Chain Rule

3 Differentiability


DefinitionIf z = f (x, y) is differentiable at (x0, y0), then the total differential of z at(x0, y0) is given by

dz = fx(x0, y0)dx+ fy (xo , y0)dy

ExampleCompute dz for the following:

1 z = x2+ y22 z = tan1(xy)

RemarkNote that dz by itself makes sense, but z by itself makes no sense.

AnnouncementsRecall Class Policy:

1 Missed midterm, prefinal or final exam, with valid excuse: SpecialExam (Harder than the usual)

2 Missed Chapter Quiz, with valid excuse: 1st missed chapter quiz willnot be included in total, succeeding missed chapter quizzes grade of 0

Implicit DifferentiationThe Chain RuleDifferentiabilityThe Total Differential

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M38 Lec 021114