3_7 - James Stewart I

a= f b= f/

c= f//

a b=0 b

c=0 c f f/

c a b 0

d c 0 d/=c c x<0 b

x<0 c/=b b R x=0

a b/=a d= f c= f

/b= f

//

a= f///

a a

c b 0 a=0 b

b b/=a

c

a 0 a 0 b

b/=a b 0 c c

/=b d

c b a

f (x)=x5+6x

27x f

/(x)=5x

4+12x 7 f

//(x)=20x

3+12

f (t)=t8

7t6+2t

4f

/(t)=8t

742t

5+8t

3f

//(t)=56t

6210t

4+24t

2

y=cos 2 y/= 2sin 2 y

//= 4cos 2

y= sin y/= cos +sin y

//= ( sin )+cos 1+cos =2cos sin

F(t)=(1 7t)6

F/(t)=6(1 7t)

5( 7)= 42(1 7t)

5F

//(t)= 42 5(1 7t)

4( 7)=1470(1 7t)

4

g(x)=2x+1

x 1g

/(x)=

(x 1)(2) (2x+1)(1)

(x 1)2

=2x 2 2x 1

(x 1)2

=3

(x 1)2

3(x 1)2

g//(x)= 3( 2)(x 1)

3=6(x 1)

3 6

(x 1)3

h(u)=1 4u

1+3uh

/(u)=

(1+3u)( 4) (1 4u)(3)

(1+3u)2

=4 12u 3+12u

(1+3u)2

=7

(1+3u)2

7(1+3u)2

h//(u)= 7( 2)(1+3u)

3(3)=42(1+3u)

3

1. , , . We can see this because where has a horizontal tangent, , and where

has a horizontal tangent, . We can immediately see that can be neither nor , since at the

points where has a horizontal tangent, neither nor is equal to .

2. Where has horizontal tangents, only is , so . has negative tangents for and is the

only graph that is negative for , so . has positive tangents on (except at ), and the

only graph that is positive on the same domain is , so . We conclude that , ,

and .

3. We can immediately see that is the graph of the acceleration function, since at the points where

has a horizontal tangent, neither nor is equal to . Next, we note that at the point where has

a horizontal tangent, so must be the graph of the velocity function, and hence, . We conclude

that is the graph of the position function.

4. must be the jerk since none of the graphs are at its high and low points. is where has a

maximum, so . is where has a maximum, so . We conclude that is the position

function, is the velocity, is the acceleration, and is the jerk.

5.

6.

7.

8.

9.

10. or

or

11. or

or

1

Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives

42

(1+3u)3

H(s) =a s+b

s=as

1/2+bs

1/2

H/(s) =a

1

2s

1/2+b

1

2s

3/2=

1

2as

1/2 1

2bs

3/2

H//(s) =

1

2a

1

2s

3/2 1

2b

3

2s

5/2=

1

4as

3/2+

3

4bs

5/2

h(x)= x2+1 h

/(x)=

1

2(x

2+1)

1/2(2x)=

x

x2+1

h//(x)=

x2+1 1 x

1

2(x

2+1)

1/2(2x)

x2+1( )

2=

(x2+1)

1/2[(x

2+1) x

2]

(x2+1)

1=

1

(x2+1)

3/2

y=xecx

y/=x e

cxc+e

cx1=e

cx(cx+1) y

//=e

cx(c)+(cx+1)e

cxc=ce

cx(1+cx+1)=ce

cx(cx+2)

y=(x3+1)

2/3y

/=

2

3(x

3+1)

1/3(3x

2)=2x

2(x

3+1)

1/3

y//=2x

2 1

3(x

3+1)

4/3(3x

2)+(x

3+1)

1/3(4x)=4x(x

3+1)

1/32x

4(x

3+1)

4/3

y =4x

x+1

y/

=

x+1 4 4x1

2(x+1)

1/2

( x+1 )2

=4 x+1 2x/ x+1

x+1=

4(x+1) 2x

(x+1)3/2

=2x+4

(x+1)3/2

y//

=

(x+1)3/2

2 (2x+4)3

2(x+1)

1/2

[(x+1)3/2

]2

=(x+1)

1/2[2(x+1) 3(x+2)]

(x+1)3

=2x+2 3x 6

(x+1)5/2

=x 4

(x+1)5/2

H(t)=tan 3t

12.

13.

14.

15.

16.

17.

2


H/(t)=3sec

23t H

//(t)=2 3sec 3t

d

dt(sec 3t)=6sec 3t (3sec 3t tan 3t)=18sec

23t tan 3t

g(s)=s2cos s g

/(s)=2s cos s s

2sin s

g//(s)=2cos s 2s sin s 2s sin s s

2cos s=(2 s

2)cos s 4s sin s

g(t)=t3e

5tg

/(t)=t

3e

5t5+e

5t3t

2=t

2e

5t(5t+3)

g//(t) =(2t)e

5t(5t+3)+t

2(5e

5t)(5t+3)+t

2e

5t(5)

=te5t

2(5t+3)+5t(5t+3)+5t =te5t

(25t2+30t+6)

h(x)=tan1(x

2) h

/(x)=

1

1+(x2)2

2x=2x

1+x4

h//(x)=

(1+x4)(2) (2x)(4x

3)

(1+x4)2

=2 6x

4

(1+x4)2

f (x)=2cos x+sin2x f

/(x)=2( sin x)+2sin x(cos x)=sin 2x 2sin x

f//(x)=2cos 2x 2cos x=2(cos 2x cos x)

f f/(x)=0 f

/

f//(x)=0

f (x)=ex

x3

f/(x)=e

x3x

2f

//(x)=e

x6x

f f/

f/

f

f/

f f/

f//.

18.

19.

20.

21. (a)

(b)

We can see that our answers are plausible, since has horizontal tangents where , and

has horizontal tangents where .

22. (a)

(b)

The graphs seem reasonable since has horizontal tangents where is zero, is positive where

is increasing, and is negative where is decreasing; and the same relationships exist between

and

3


y= 2x+3=(2x+3)1/2

y/=

1

2(2x+3)

1/22=(2x+3)

1/2y

//=

1

2(2x+3)

3/22= (2x+3)

3/2

y///

=3

2(2x+3)

5/22=3(2x+3)

5/2

y=x

2x 1y

/=

(2x 1)(1) x(2)

(2x 1)2

=1

(2x 1)2

1(2x 1)2

y//= 1( 2)(2x 1)

3(2)=4(2x 1)

3

y///

=4( 3)(2x 1)4(2)= 24(2x 1)

424/(2x 1)

4

f (t)=tcos t f/(t)=t( sin t)+cos t 1 f

//(t)=t( cos t) sin t 1 sin t

f///

(t)=tsin t cos t 1 cos t cos t=tsin t 3cos t f///

(0)=0 3= 3

g(x)= 5 2x g/(x)=

1

2(5 2x)

1/2( 2)= (5 2x)

1/2g

//(x)=

1

2(5 2x)

3/2( 2)= (5 2x)

3/2

g///

(x)=3

2(5 2x)

5/2( 2)= 3(5 2x)

5/2g

///(2)= 3(1)

5/2= 3

f ( )=cot f/( )= csc

2f

//( )= 2csc ( csc cot )=2csc

2cot

f///

( )=2( 2csc2

cot )cot +2csc2

( csc2

)= 2csc2

(2cot2

+csc2

)

f///

6= 2(2)

2[2( 3 )

2+(2)

2]= 80

g(x)=sec x g/(x)=sec x tan x

g//(x)=sec x sec

2x+tan x(sec x tan x)=sec

3x+sec x tan

2x=sec

3x+sec x(sec

2x 1)=2sec

3x sec x

g///

(x)=6sec2x(sec x tan x) sec x tan x=sec x tan x(6sec

2x 1) g

///

4= 2 (1)(6 2 1)=11 2

9x2+y

2=9 18x+2yy

/=0 2yy

/= 18x y

/= 9x/y

y//= 9

y 1 x y/

y2

= 9y x( 9x/y)

y2

= 9y

2+9x

2

y3

= 99

y3

9x2+y

2=9] y

//= 81/y

3

23.

24. or

or

25.

, so .

26.

, so .

27.

28.

29.

[ since x and y must satisfy the

original equation, . Thus, .

30.

4


x + y=11

2 x+

y/

2 y=0 y

/=

y

x

y//

=

x1

(2 y )y

/y

1

(2 x )

x=

x1

y

y

xy

1

x

2x=

1+y

x

2x

=x + y

2x x=

1

2x xx y x + y=1

x3+y

3=1 3x

2+3y

2y

/=0 y

/=

x2

y2

y//=

y2(2x) x

22yy

/

(y2)2

=

2xy2

2x2y

x2

y2

y4

=2xy

4+2x

4y

y6

=2xy(y

3+x

3)

y6

=2x

y5

x y x3+y

3=1

x4+y

4=a

44x

3+4y

3y

/=0 4y

3y

/= 4x

3y

/= x

3/y

3

y//=

y3

3x2

x3

3y2y

/

(y3)2

= 3x2y

2

y xx

3

y3

y6

= 3x2 y

4+x

4

y4y

3= 3x

2 a4

y7

=3a

4x

2

y7

f (x)=xn

f/(x)=nx

n 1f

//(x)=n(n 1)x

n 2f

n( )(x)=n(n 1)(n 2) 2 1x

n n=n!

f (x)=1

5x 1=(5x 1)

1f

/(x)= 1(5x 1)

25 f

//(x)=( 1)( 2)(5x 1)

35

2

f///

(x)=( 1)( 2)( 3)(5x 1)4

53

f(n)

(x)=( 1)nn!5

n(5x 1)

(n+1)

f (x)=e2x

f/(x)=2e

2xf

//(x)=2 2e

2x=2

2e

2x

f///

(x)=22

2e2x

=23e

2xf

(n)(x)=2

ne

2x

f (x)= x =x1/2

f/(x)=

1

2x

1/2

f//(x)=

1

2

1

2x

3/2f

///(x)=

1

2

1

2

3

2x

5/2

since and must satisfy the original equation, .

31.

,

since and must satisfy the original equation, .

32.

33.

34.

35.

36.

5


f4( )

(x)=1

2

1

2

3

2

5

2x

7/2=

1 3 5

24

x7/2

f5( )

(x)=1

2

1

2

3

2

5

2

7

2x

9/2=

1 3 5 7

25

x9/2

fn( )

(x)=1

2

1

2

3

2

1

2n+1 x

2n 1( ) /2=( 1)

n 1 1 3 5 (2n 3)

2n

x2n 1( ) /2

f (x)=1/(3x3)=

1

3x

3f

/(x)=

1

3( 3)x

4f

//(x)=

1

3( 3)( 4)x

5

f///

(x)=1

3( 3)( 4)( 5)x

6

fn( )

(x)=1

3( 3)( 4) [ (n+2)]x

n+3( )=

( 1)n

3 4 5 (n+2)

3xn+3

2

2=

( 1)n(n+2)!

6xn+3

Dsin x=cos x D2sin x= sin x D

3sin x= cos x D

4sin x=sin x sin x

74=4(18)+2 D74

sin x=D2sin x= sin x

f (x)=cos x Df (2x)=2 f/(2x) D

2f (2x)=2

2f

//(2x) D

3f (2x)=2

3f

///(2x) . . .

D(n)

f (2x)=2n

f(n)

(2x) cos x 103=4(25)+3

f(103)

(x)= f(3)

(x)=sin x D103

cos 2x=2103

f(103)

(2x)=2103

sin 2x

f (x)=xex

f/(x)=x( e

x)+e

x=(1 x)e

xf

//(x)=(1 x)( e

x)+e

x( 1)=(x 2)e

x

f///

(x)=(x 2)( ex)+e

x=(3 x)e

xf

(4)(x)=(3 x)( e

x)+e

x( 1)=(x 4)e

x

f(n)

(x)=( 1)n(x n)e

x

D1000

xex=(x 1000)e

x.

s= f (t) t=0 1 2 3 4 5

v= f/(t)

t=2 a= f//(2) f

/t=2

a(2)= f//(2)=v

/(2)

27

3=9 /

2

37.

38. . The derivatives of occur in

a cycle of four. Since , we have .

39. Let . Then , , , ,

. Since the derivatives of occur in a cycle of four, and since ,

we have and .

40.

.

So

41. By measuring the slope of the graph of at , , , , , and , and using the method

of Example 1 in Section 3.2, we plot the graph of the velocity function in the first figure. The

acceleration when s is , the slope of the tangent line to the graph of when . We

estimate the slope of this tangent line to be ft s . Similar measurements

enable us to graph the acceleration function in the second figure.

6


t=10 0 t=10

t=10 a a 10 /2

20 t=10 10/20= 0.5 (ft/s2). / /

3

s=2t3

15t2+36t+2 v(t)=s

/(t)=6t

230t+36 a(t)=v

/(t)=12t 30

a(1)=12 1 30= 18 /2

v(t)=6(t2

5t+6)=6(t 2)(t 3)=0 t=2 3 a(2)=24 30= 6 /2

a(3)=36 30=6 /2.

s=2t3

3t2

12t v(t)=s/(t)=6t

26t 12 a(t)=v

/(t)=12t 6

a(1)=12 1 6=6 /2

v(t)=6(t2

t 2)=6(t+1)(t 2)=0 t= 1 2 t 0 t 1

a(2)=24 6=18 /2.

s=sin6

t +cos6

t 0 t 2

v(t)=s/(t)=cos

6t

6sin

6t

6=

6cos

6t sin

6t

a(t)=v/(t)=

6sin

6t

6cos

6t

6=

2

36sin

6t +cos

6t

a(1)=

2

36sin

61 +cos

61 =

2

36

1

2+

3

2=

2

72(1+ 3) 0.3745 /

2

v(t)=0 0 t 2 cos6

t =sin6

t

42. (a) Since we estimate the velocity to be a maximum at , the acceleration is at .

(b) Drawing a tangent line at on the graph of , appears to decrease by ft s over a

period of s. So at s, the jerk is approximately s or ft s .

43. (a)

(b) m s

(c) when or and m s , m s

44. (a)

(b) m s

(c) when or . Since , and

m s

45. (a) , .

(b) m s

(c) for

7


1=

sin6

t

cos6

t

tan6

t =16

t=tan11 t=

6

4=

3

2=1.5

a(3

2)=

2

36sin

6

3

2+cos

6

3

2=

2

36

2

2+

2

2=

2

362 0.3877 /

2

s=2t3

7t2+4t+1 v(t)=s

/(t)=6t

214t+4 a(t)=v

/(t)=12t 14

a(1)=12 14= 2 /2

v(t)=2 3t2

7t+2( )=2(3t 1)(t 2)=0 t=1

32 a

1

3=12

1

314= 10 /

2

a(2)=12(2) 14=10 /2

s(t)=t4

4t3+2 v(t)=s

/(t)=4t

312t

2a(t)=v

/(t)=12t

224t=12t(t 2)=0 t=0 2

s(0)=2 v(0)=0 / s(2)= 14 v(2)= 16 /

s(t)=2t3

9t2

v(t)=s/(t)=6t

218t a(t)=v

/t( )=12t 18=0 t=1.5

s(1.5)= 13.5 v(1.5)= 13.5 /

s= f (t)=t3

12t2+36t t 0 v(t)= f

/(t)=3t

224t+36

a(t)=v/(t)=6t 24 a(3)=6(3) 24= 6 m/s( ) / /

2

v a 2<t<4

t>6 v a 0 t<2 4<t<6

x(t)=t

1+t2

v(t)=x/(t)=

(1+t2)(1) t(2t)

(1+t2)2

=1 t

2

(1+t2)2

a(t)=v/(t)=

2t (t2

3)

(1+t2)3

a(t)=0 2t (t2

3)=0 t=0 3

s. Thus,

m s .

46. (a)

(b) m s

(c) when or and m s ,

m s .

47. (a) when or .

(b) m, m s, m, m s

48. (a) when .

(b) m, m s

49. (a) , .

. s or m s .

(b)

(c) The particle is speeding up when and have the same sign. This occurs when and when

. It is slowing down when and have opposite signs; that is, when and when .

50. (a) .

. or

(b)

8


v a 1<t< 3

v a 0<t<1 t> 3

y(t)=Asin t v(t)=y/(t)=A cos t a(t)=v

/(t)= A

2sin t

a(t)= A2sin t=

2(Asin t)=

2y(t) a(t) y(t)

= v(t) =A cos t cos t= 1 cos t= 1

sin t=0 a(t)= A2sin t= A

2(0)=0

a(t)=dv

dt=

dv

ds

ds

dt=

dv

dsv(t)=v(t)

dv

dsdv/dt

dv/ds

P(x)=ax2+bx+c P

/(x)=2ax+b P

//(x)=2a P

//(2)=2 2a=2 a=1

P/(2)=3 2(1)(2)+b=3 4+b=3 b= 1

P(2)=5 1(2)2+( 1)(2)+c=5 2+c=5 c=3 P(x)=x

2x+3

Q(x)=ax3+bx

2+cx+d Q

/(x)=3ax

2+2bx+c Q

//(x)=6ax+2b Q

///(x)=6a

Q 1( )=a+b+c+d=1 Q/

1( )=3a+2b+c=3 Q//(1)=6a+2b=6 Q

///(1)=6a=12

a b c d a=2 b= 3 c=3 d= 1 Q(x)=2x3

3x2+3x 1

y=Asin x+Bcos x y/=Acos x Bsin x y

//= Asin x Bcos x y

//+y

/2y=sin x

( 3A B)sin x+(A 3B)cos x=1sin x 3A B=1 A 3B=0 A

B B=1

10A=

3

10

y=Ax2+Bx+C y

/=2Ax+B y

//=2A

y//+y

/2y=x

2

(2A)+(2Ax+B) 2(Ax2+Bx+C) = x

2

2A+2Ax+B 2Ax2

2Bx 2C = x2

(c) and have the same sign and the particle is speeding up when . The particle is slowing

down and and have opposite signs when and when .

51. (a)

(b) , so is proportional to .

(c) speed is a maximum when . But when , we have

, and .

52. By the Chain Rule, . The derivative is the rate of change

of the velocity with respect to time (in other words, the acceleration) whereas the derivative is

the rate of change of the velocity with respect to the displacement.

53. Let . Then and . .

.

. So .

54. Let . Then , and . Thus,

, , and . Solving these four

equations in four unknowns , , and we get , , and , so

.

55. . Substituting into

gives us , so we must have and . Solving for

and , we add the first equation to three times the second to get and .

56. . We substitute these expressions into the equation

to get

9


( 2A)x2+(2A 2B)x+(2A+B 2C) = (1)x

2+(0)x+(0)

x2

2A=1 A=1

22A 2B=0

A=B=1

22A+B 2C=0 1

1

22C=0 C=

3

4

y=erx

y/=re

rxy

//=r

2e

rx, y

//+5y

/6y=r

2e

rx+5re

rx6e

rx=e

rx(r

2+5r 6)=e

rx(r+6)(r 1)=0

(r+6)(r 1)=0 r=1 6.

y=ex

y/= e

xy

//=

2e

x. y+y

/=y

//e

x+ e

x=

2e

xe

x(

21)=0

=1 5

2, e

x0.

f (x)=xg(x2) f

/(x)=x g

/(x

2) 2x+g(x

2) 1=g(x

2)+2x

2g

/(x

2)

f//(x)=g

/(x

2) 2x+2x

2g

//(x

2) 2x+g

/(x

2) 4x=6xg

/(x

2)+4x

3g

//(x

2)

f (x)=g(x)

xf

/(x)=

xg/(x) g(x)

x2

f//(x)=

x2[g

/(x)+xg

//(x) g

/(x)] 2x[xg

/(x) g(x)]

x4

=x

2g

//(x) 2xg

/(x)+2g(x)

x3

f (x) =g( x ) f/(x)=g

/( x )

1

2x

1/2=

g/( x )

2 x

f//(x) =

2 x g//( x )

1

2x

1/2g

/( x ) 2

1

2x

1/2

(2 x )2

=x

1/2[ x g

//( x ) g

/( x )]

4x

=x g

//( x ) g

/( x )

4x x

The coefficients of on each side must be equal, so . Similarly,

and .

57. so

or

58. Thus,

since

59.

60.

61.

62.

10


f (x)=3x5

10x3+5 f

/(x)=15x

430x

2f

//(x)=60x

360x=60x(x

21)=60x(x+1)(x 1)

f//(x)>0 1<x<0 x>1 f

f//(x)<0 x< 1 0<x<1 f

f (x) =1

x2+x

f/(x)=

(2x+1)

(x2+x)

2

f//(x) =

(x2+x)

2( 2)+(2x+1)(2)(x

2+x)(2x+1)

(x2+x)

4=

2(3x2+3x+1)

(x2+x)

3

f///

(x)=(x

2+x)

3(2)(6x+3) 2(3x

2+3x+1)(3)(x

2+x)

2(2x+1)

(x2+x)

6

=6(4x

3+6x

2+4x+1)

(x2+x)

4

f(4)

(x) =(x

2+x)

4( 6)(12x

2+12x+4)+6(4x

3+6x

2+4x+1)(4)(x

2+x)

3(2x+1)

(x2+x)

8

=24(5x

4+10x

3+10x

2+5x+1)

(x2+x)

5

f(5)

(x) =?

f (x)=1

x(x+1)=

1

x

1

x+1f

/(x)= x

2+(x+1)

2f

//(x)=2x

32(x+1)

3

f///

(x)=( 3)(2)x4+(3)(2)(x+1)

4f

n( )(x)=( 1)

nn! x

n+1( )(x+1)

n+1( )

f (x)=7x+17

2x2

7x 4

f///

(x)=6(56x

4+544x

32184x

2+6184x 6139)

(2x2

7x 4)4

f (x)=7x+17

2x2

7x 4

=3

2x+1+

5

x 4

f///

(x)=144

(2x+1)4

30

(x 4)4

f (x)=x2e

x, f

/(x)=x

2e

x+e

x(2x) =e

x(x

2+2x).

So when or , and on these intervals the graph of lies above its tangent lines;

and when or , and on these intervals the graph of lies below its tangent lines.

63. (a)

(b)

64. (a) For , a CAS gives us

(b) Using a CAS we get . Now we differentiate three times to obtain

.

65.

For Similarly, we have

11


f//(x) =e

x(x

2+4x+2)

f///

(x) =ex(x

2+6x+6)

f(4)

(x) =ex(x

2+8x+12)

f(5)

(x) =ex(x

2+10x+20)

x 2

0=1 0, 2=2 1, 6=3 2, 12=4 3, 20=5 4.

f(n)

(x)=ex

x2+2nx+n(n 1) .

Sn

f(n)

(x)=ex

x2+2nx+n(n 1) . S

1

f/(x)=e

x(x

2+2x).

Sk

f(k)

(x)=ex

x2+2kx+k(k 1) .

f(k+1)

(x) =d

dxf

(k)(x) =e

x(2x+2k)+ x

2+2kx+k(k 1) e

x

=ex

x2+(2k+2)x+(k

2+k) =e

xx

2+2(k+1)x+(k+1)k

Sk+1

Sn

n; f(n)

(x)=ex

x2+2nx+n(n 1)

n.

F= fg F/= f

/g+ fg

/

F//=( f

//g+ f

/g

/)+( f

/g

/+ fg

//)= f

//g+2 f

/g

/+ fg

//

F///

= f///

g+ f//

g/+2( f

//g

/+ f

/g

//)+ f

/g

//+ fg

///= f

///g+3 f

//g

/+3 f

/g

//+ fg

///

F4( )

= f4( )

g+ f///

g/+3( f

///g

/+ f

//g

//)+3( f

//g

//+ f

/g

///)+ f

/g

///+ fg

4( )

= f4( )

g+4 f///

g/+6 f

//g

//+4 f

/g

///+ fg

4( )

Fn( )

= fn( )

g+nfn 1( )

g/+ n

2f

n 2( )g

//+ + n

kf

n k( )g

k( )+ +nf

/g

n 1( )+ fg

n( )

nk

=n!

k!(n k)!=

n(n 1)(n 2) (n k+1)

k!

It appears that the coefficient of in the quadratic term increases by with each differentiation. The

pattern for the constant terms seems to be So a reasonable

guess is that

Proof: Let be the statement that 1. is true because

2. Assume that is true; that is, Then

This shows that is true.

3. Therefore, by mathematical induction, is true for all that is, for

every positive integer

66. (a) Use the Product Rule repeatedly:

.

(b)

(c) By analogy with the Binomial Theorem, we make the guess:

where .

67. The Chain Rule says that

12


dy

dx=

dy

du

du

dx

d2y

dx2

=d

dx

dy

dx=

d

dx

dy

du

du

dx=

d

dx

dy

du

du

dx+

dy

du

d

dx

du

dx

=d

du

dy

du

du

dx

du

dx+

dy

du

d2u

dx2

=d

2y

du2

du

dx

2

+dy

du

d2u

dx2

d2y

dx2

=d

2y

du2

du

dx

2

+dy

du

d2u

dx2

d3y

dx3

=d

dx

d2y

dx2

=d

dx

d2y

du2

du

dx

2

+d

dx

dy

du

d2u

dx2

=d

dx

d2y

du2

du

dx

2

+d

dx

du

dx

2 d2y

du2

+d

dx

dy

du

d2u

dx2

+d

dx

d2u

dx2

dy

du

=d

du

d2y

du2

du

dx

du

dx

2

+2du

dx

d2u

dx2

d2y

du2

+d

du

dy

du

du

dx

d2u

dx2

+d

3u

dx3

dy

du

=d

3y

du3

du

dx

3

+3du

dx

d2u

dx2

d2y

du2

+dy

du

d3u

dx3

n n f(n)

f

f/

f//

f///

. . . f(p 1)

. f(p)

= f p f f/

f//

f///

. . . f(p 1)

f. n=1 k k 1 f k

f(k)

S={ f f/

f//

. . . f(p 1)

} f p S

f(k+1)

S f(k+1)

{ f/

f//

f///

. . . f(p)

} S f(p)

= f n=1

n=k n=k+1

n.

, so

[Product Rule]

68. From Exercise 65,

69. We will show that for each positive integer , the th derivative exists and equals one of ,

, , , , Since , the first derivatives of are , , , , ,

and In particular, our statement is true for . Suppose that is an integer, , for which is

times differentiable with in the set

, , , , . Since is times differentiable, every member of is differentiable,

so exists and equals the derivative of some member of . Thus, is in the set , ,

, , , which equals since . We have shown that the statement is true for and

that its truth for implies its truth for . By mathematical induction, the statement is true for

all positive integers

13


Documents

3_7 - James Stewart I