Matematica

Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions

e 1 =1 . 1. (a) e is the number such that lim h 0 h (b) x 0.001 0.0001 0.001 0.0001 (2.7 1)/x 0.9928 0.9932 0.9937 0.9933
x

h

x 0.001 0.0001 0.001 0.0001

(2.8 1)/x 1.0291 1.0296 1.0301 1.0297
h h

x

2.7 1 2.8 1 From the tables (to two decimal places), lim =0.99and lim =1.03 . Since h h h 0 h 0 0.990 , so coth y= x +1 . If x0 .) dx sech ytanh y 2 2 sech y 1 sech y x 1 x dy 1 1 1 2 dy 1 (e) Let y=coth x. Then coth y=x csch y =1 = = = by Exercise 2 2 2 dx dx csch y 1 coth y 1 x 13. 30. f (x)=tanh 4x 31. f (x)=xcosh x 32. g(x)=sinh x 33. h(x)=sinh (x )
2 2

f (x)=4sech 4x f (x)=x(cosh x) +(cosh x)(x) =xsinh x+cosh x g (x)=2sinh xcosh x h (x)=cosh (x )2x=2xcosh (x ) F (x)=sinh xsech x+tanh xcosh x
/ 2 / 2 2 / / / /

/

2

34. F(x)=sinh xtanh x 35. G(x)= G (x)
/

=

1 cosh x 1+cosh x (1+cosh x)( sinh x) (1 cosh x)(sinh x) (1+cosh x)
2

7

Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions

=

sinh x sinh xcosh x sinh x+sinh xcosh x (1+cosh x)
2

=

2sinh x (1+cosh x)
2

36. f (t)=esech t

t

f (t)=e ( sech ttanh t)+(sech t)e =e sech t (1 tanh t)
2 / 2 2

/

t

t

t

37. h(t)=coth

1+t

h (t)= csch

1+t

1 2 (1+t ) 2

1/2

(2t)=

tcsch

2

1+t
2

2

1+t

38. f (t)=ln (sinh t) 39. H(t)=tanh (e ) 40. y=sinh (cosh x) 41. y=e
cosh 3x t

f (t)=
/

/

1 cosh t=coth t sinh t
2 t t t 2 t

H (t)=sech (e ) e =e sech (e ) y =cosh (cosh x)sinh x
cosh 3x /

y =e
1

/

sinh 3x 3=3e
2

cosh 3x

sinh 3x
1

42. y=x sinh (2x)

2

y =x

/

1 1+(2x)
2

2+sinh (2x) 2x=2x

x 1+4x
2

+sinh (2x)

1

43. y=tanh

1

x

y =

/

1 1 ( x)
2

1 x 2
1

1/2

=

1 2 x (1 x)

44. y=xtanh x+ln y =tanh x+
/ 1

1

1 x =xtanh x+ + 1 2
2

2

1 2 ln (1 x ) 2
1

x 1 x
2

1 1 x
2

(2x)=tanh x

45. y=xsinh (x/3) 9+x x / 1 1/3 y =sinh +x 3 2 1+(x/3) 46. y=sech
1

1

2x 2 9+x
2

=sinh

1

x 3

+

x 9+x
2

x 9+x
2

=sinh

1

x 3

1 x

2

8

Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions

y =

/

1 1 x
2

2x
2

=
2

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

1 (1 x ) 2 x +1
2

1 x 1

47. y=coth

1y =

/

1 (x +1) 2

2

= x

1 x +1
2

48. For y=acosh (x/a) with a>0 , we have the y intercept equal to a. As a increases, the graph flattens. 49. (a) y=20cosh (x/20) 15 x=7, we have y (7)=sinh (b) If
/

y =20sinh (x/20) 0.3572 .

/

1 =sinh (x/20) . Since the right pole is positioned at 20

7 20

is the angle between the tangent line and the x axis, then tan = slope ofthe line =sinh
1

7 20

, so =tan =90

sinh

7 20

0.343 rad

19.66 . Thus, the angle between the line and the pole is

70.34 . T gx cosh g T

50. We differentiate the function twice, then substitute into the differential equation: y= dy T = sinh dx g gx T g gx =sinh T T d y dx d y dx RHS= g T 1+ dy dx
2 2 2 2 2

=cosh

gx T

g g gx = cosh . T T T

We evaluate the twosides separately: LHS= g T
2

=

g gx cosh , T T

=

1+sinh
/

gx g gx = by the identity proved in Example 1(a). T T T

51. (a) y=Asinh mx+Bcosh mx
// 2 2

y =mAcosh mx+mBsinh mx
2 2 //

y =m Asinh mx+m Bcosh mx=m (Asinh mx+Bcosh mx)=m y (b) From part (a), a solution of y =9y is y(x)=Asinh 3x+Bcosh 3x. So 4=y(0)=Asinh 0+Bcosh 0=B,
9

Stewart Calculus ET 5e 0534393217;3.Differentiation Rules; 3.9 Hyperbolic Functions

so B= 4. Now y (x)=3Acosh 3x 12sinh 3x sinh x e
x

/

6=y (0)=3A 1 0 1 = 2 2
/

/

A=2, so y=2sinh 3x 4cosh 3x .

52. lim
x

=lim
x

e e 2e
x

x

x

1 e =lim 2 x

2x

=

53. The tangent to y=cosh x has slope 1 when y =sinh x=1
2

x=sinh 1=ln (1+ 2 ) , by Equation 3.

1

Since sinh x=1 and y=cosh x= 1+sinh x , we...