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### Transcript of 1.1 Homework Solutions - Kevin Quattrin, .! 5! 1.3 Homework Solutions 1. Find the equation of the

• 1

1.1 Homework Solutions 1. f (x)= x

2 +3x 4 f (x)= 2x+3

3. y = x

23 dydx =

23x

53

5.

v(r)= 4

3r3 v(r)= 4r2

7. y = x

2 + 4x+3x

= x3

2 + 4x1

2 +3x1

2

dydx =

32x

12 + 2x

12 3

2x3

2 = 3 x2

+ 2x 3

2 x3

9. z = Ay10 + Bey = Ay10 + Bey

z = 10Ay11 + Bey 11. If f (x) = 3x

5 5x3 + 3 , find f '(x). f x( ) = 45x14 15x2 You can see that when f is decreasing the graph of f is below the x - axis, i.e. f is negative - if f is increasing the graph of f is above the x - axis.

13. y = e x = ex12 dydx = e

x12 12 x

12 = ex12

2x12= e

x

2 x y = e x

• 2

15. Given the following table of values, find the indicated derivatives.

x f (x) f (x)2 1 78 5 3

a. g x( ) = f x( )

3

g x( ) = 3 f x( ) 2 f x( ) g 2( ) = 3 f 2( )

2 f 2( ) = 312 7 = 21

b. h x( ) = f x3( ) h x( ) = f x3( )3x2 h 2( ) = f 23( )322 =12 f 8( ) =12 3= 36

17. If

f (x) = x3 + 2x( )37 , find f '(x) .

f x( ) = 37 x3 + 2x( )36 3x2 + 2( ) = 37 3x2 + 2( ) x3 + 2x( )36

19. f x( ) = 4 49 x

2 ; find f ' 5( )

f x( ) = 4 49 x2 = 4 49 x

2

12 f x( ) = 12 4

49 x

2

12 89 x

= 4x

9 4 49 x2

21. y = x3 2x7 ; find dy

dx

y = x3 2x7 = x3 2x( )17 f x( ) = 17 x3 2x( )67 3x2 2( ) = 3x

2 2

7 x3 2x( )67

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1.2 Homework Solutions 1. sin 4y x=

y = 4cos4x 3. y = a

3 + cos3 x y = 3cos2 x sin x( ) = 3cos2 xsin x

5. f (t) = 1+ tant

3

f t( ) = 13 1+ tant( )23 sec2 t = sec

2 t1+ tant( )23

7.

y = cos a3+x3( )

y = sin a3 + x3( )3x2 = 3x2 sin a3 + x3( ) 9.

f (x) = cos ln x( )

f x( ) = sin ln x( )1x = sin ln x( )

x

11.

f (x) = log10 2+ sin x( ) f x( ) = 12+ sin x

1ln10 cosx =

cosxln10 2+ sin x( )

13.

y = sin1 ex( )

y = 11 e2x

ex = ex

1 e2x

15.

y = sin1 2x +1( )

y = 11 2x+1( )2

2 = 21 4x2 + 4x+1( )

= 2 4x2 4x( )12 = 1x2 x( )12

• 4

17. y = csc1 x2 +1( ) y = 1

x2 +1( ) x2 +1( )2 12x = 2x

x2 +1( ) x2 +1( )2 1

• 5

1.3 Homework Solutions 1. Find the equation of the tangent line to f (x) = x5 5x +1 at x = 2 and use it to get an approximate value of f (-1.9).

f 2( ) = 21 f x( ) = 5x4 5 f 2( ) = 75 y+ 21= 75 x+ 2( )y = 21+ 75 x+ 2( ) y x=1.9 = 13.5

f 1.9( ) 13.5 3. Find an equation of the line tangent to the curve y = x

4 + 2ex at the point

(0,2) .

y = 4x3 + 2ex y x=0 = 2 Tangent Line: y 2 = 2 x 0( )

5. Find the equation of the tangent line to cosy x x= + at the point

0,1( ) .

y =1 sin x y x=0 =1 Tangent Line: y1=1 x 0( )

7. Find the equation of the line tangent to y = 2 x + cos 4x( ) when x =

2

.

y = 2 4sin 4x( ) y x= 2 =2 4sin 2( ) =

2

yx= 2

= 2 2 + cos 2( ) = 2

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9. Use Eulers Method with 4 equal step sizes to find an approximation for

f 1.4( ) , given that

f x( ) = ln 2x 1( ) and f 1( ) = 0 .

f 1.4( ) 0.635

11. Given the differential equation,

dydx

= y8

6 y( ) where y = f t( ) and f 0( ) = 8 , use Eulers Method with two steps of equal size to approximate

f 1( ) .

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13. Let y = f x( ) be the solution to the differential equation dy x ydx = + with the

initial condition f 1( ) = 2 . What is the approximation for f 2( ) if Eulers method is

used, starting at x = 1 with a step size of 0.5? (A) 3 (B) 5 (C) 6 (D) 10 (E) 12

• 8

1.4 Homework Solutions 1. 3 cosy t t=

y = t 3 sint( )+ cost 3t2 = t2 3cost t sint( )

3. y = tan x 1

sec x

y =secx sec2 x tan x1( )secx tan x

sec2 x =sec2 x tan x1( )tan x

secx =sec2 x tan2 x+ tan x

secx Recall 1= sec2 x tan2 x , so substitute into numerator:

y = 1+ tan xsecx

5. y = xe

x2 y = x ex

2 2x( )+ ex2 = ex2 1 2x2 7. y = e

xcosx y = excosx x sin x+ cosx = e

xcosx cosx xsin x

9. y = xsin 1

x

y = x cos 1x 1x2

+ sin 1x = 1x cos

1x + sin

1x

11. f (x) = x ln x

f x( ) = x 12 ln x( )12 1x + ln x( )

12 = 1+ 2ln x2 ln x( )12

• 9

13. Find the equation of the line tangent to y = x2ex at the point

1, 1e( ) .

y = x2 ex 1( )+ ex 2x m = y x=1 = 1e +2e =

1e

y 1e =1e x1( )

15. h(t) = 1+ x

2

1 x2

17

, find h '(t)

h t( ) =17 1+ t2

1 t2

16

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

1 t2( )2=17 1+ t2( )16 2t 2t 3 + 2t + 2t 3

1 t2( )18

h t( ) =68t 1+ t 22( )161 t 2( )18

17. f (x) = xsin 2x( ) + tan4 x7( )

5, find f ' x( ) .

f x( ) = 5 xsin 2x( )+ tan4 x7( ) 4 xcos2x 2+ sin2x+ 4 tan3 x7( )sec2 x7( )7x6

= 5 xsin 2x( )+ tan4 x7( ) 4 2xcos2x+ sin2x+ 28x6 tan3 x7( )sec2 x7( )

19. Find the equation of the line tangent to y = e

x sin 4x( ) + 2 when x = 0 y x=0 = e

0sin0 + 2 = 3 0,3( ) y = exsin4x x cos4x 4 + sin4x y x=0 = e

0 0 cos0 4 + sin0 = 0

y 3= 0 x 3( ) y = 3

20. Find the equation of the lines tangent and normal to y = xsin 1

x

when

21.

H(x) = 1+ x2( ) tan1(x)

• 10

H x( ) = 1+ x2( ) 11+ x2 + tan1 x 2x

23. If f (x) = e

x x2 arctan x , find f ' x( ) .

f x( ) = ex x2 11+ x2 + arctan x 2x

= ex x2

1+ x2 2xarctan x

25. y = sec1 xx

y =x 1x x2 1

sec1 x

x2=

1x2 1

sec1 x

x2

• 11

1.5 Homework Solutions 1. f (x) = x

5 + 6x2 7x

f x( ) = 5x4 +12x 7f x( ) = 20x3 +12

3. y = x3 +1( )23

y = 23 x3 +1( )13 3x2 = 2x

2

x3 +1( )13

y =x3 +1( )13 4x 2x2 13 x3 +1( )

23 3x2

x3 +1( )23=x3 +1( )23 x3 +1( )4x 2x4

x3 +1( )23

= 4x4 + 4x 2x4

x3 +1( )4 3=2x x3 + 2( )x3 +1( )4 3

5. g(t) = t

3e5t

g t( ) = t 3 e5t 5+ e5t 3t2 = t2e5t 5t + 3 g t( ) = t2e5t 5+ t2 5t + 3( )e5t 5+ e5t 5t + 3( )2t

= te5t 5t + 5t 5t + 3( )+ 2 5t + 3( ) = te5t 25t2 + 30t + 6

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7. y = sin

3x

y = 3sin2 xcosx

y = 3 sin2 x sin x+ cosx 2sin xcosx = 3sin x 2cos2 x sin2 x

9. d2 dx2

5x4 + 9x3 4x2 + x 8

= ddx 20x

3 + 27x2 8x+1 = 60x2 + 54x 8

11. y = cosx2 , find y

dydx = sin x

2 2x( ) = 2xsin x2

d2ydx2

= 2x cosx2 2x( ) + sin x2 2( ) = 2 2x2 cosx2 + sin x2

13. y = sec3x , find d2ydx2

dydx = sec3x tan3x 3( ) = 3sec3x tan3xd2ydx2

= 3sec3x sec2 3x 3( )( ) + 3tan3x 3sec3x tan3x( ) = 3sec3x 3sec2 3x+ tan2 3x( )

15. f x( ) = ln x2 + 3( ) , find f x( )

f x( ) = 2xx2 + 3

f x( ) =x2 + 3( ) 2( ) 2x( ) 2x( )

x2 + 3( )2=

2x2 + 6( ) 4x2( )x2 + 3( )2

=2 x2 3( ) x2 + 3( )2

• 13

17. h x( ) = x2 + 5 , find h x( ) h x( ) = x2 + 5 = x2 + 5( )12 h x( ) = x2 + 5 = 12 x

2 + 5( )12 2x( ) = xx2 + 5( )12

h x( ) =x2 + 5( )12 1( ) x x

x2 + 5( )12x2 + 5( )1

=x2 + 5( )1 x2x2 + 5( )32

= 5

x2 + 5( )32

19. y = x2 3

x2 10, find d

2ydx2

dydx =

x2 10( ) 2x( ) x2 3( ) 2x( )x2 10( )2

=2x( ) x2 10( ) x2 3( )

x2 10( )2= 14xx2 10( )2

d2ydx2

= x2 10( )2 14( ) 14x( )2 x2 10( )1 2x( )

x2 10( )4

=14 x2 10( ) x2 10( ) 4x2( )

x2 10( )4

=14 3x2 10( )

x2 10( )3=

14 3x2 +10( )x2 10( )3

21. y = x3 + x2 7x 15

y = 3x2 + 2x 7y = 6x+ 2

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23. y = 4xx2 + 4

dydx =

x2 + 4( ) 4( ) 4x( ) 2x( )x2 + 4( )2

=4x2 16( ) 8x2( )

x2 + 4( )2= 4x

2 16x2 + 4( )2

d2ydx2

=x2 + 4( )2 8x( ) 4x2 16( )2 x2 + 4( )1 2x( )

x2 + 4( )4

=x2 + 4( ) 8x( ) 4x2 16( )2 2x( )

x2 + 4( )3

=8x3 + 32x( ) 16x3 64( )

x2 + 4( )3

=8x3 + 96x( )x2 + 4( )3

=8x x2 12( )x2 + 4( )3

25. y = x 8 x2 = x 8 x2( )12 dydx = x

12 8 x

2( )12 2x( )+ 8 x2( )12 1( )= x

2

8 x2( )12+ 8 x2( )12

=x2 + 8 x2( )18 x2( )12

= 8 2x2

8 x2( )12

• 15

d2ydx2

=

8 x2( )12 4x( ) 8 2x2( ) x28 x2( )12

8 x2( )1

=8 x2( )1 4x( ) x2( ) 8 2x2( )

8 x2( )32= 2x

4 + 4x3 + 8x2 32x

8 x2( )32

27. y = xex

dydx = xe

x 1( )+ ex 1( ) = ex x+1( )d2ydx2

= ex 1( )+ x+1( )ex 1( ) = ex x 2( )

29. y = x

x2 9

dydx =

x2 9( ) 1( ) x( ) 2x( )x2 9( )2

= x2 9

x2 9( )2

d2ydx2

=x2 9( )2 2x( ) x2 9( )2 x2 9( )1 2x( )

x2 9( )4

=x2 9( ) 2x( ) x2 9( )2 2x( )

x2 9( )3

=2x3 +18x( ) 4x3 36x( )

x2 9( )3

=2x3 + 54x( )x2 9( )3

=2x x2 + 27( )x2 9( )3

• 16

1.6 Homework Solutions 1. xy + 2x + 3x

2 = 4 Implicit Explicit

x dydx + y 1+ 2+ 6x = 0 xy = 4 2x 3x2

dydx =

6x y 2x y =

4x 2 3x = 4x

2 3= 4x2

3

dydx =4 3x2

x2

dydx =

6x y 2x =

6x 4x 2 3x

2

x = 64x2

+ 2x + 32x