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1.1 Homework Solutions 1. f (x)= x

2 +3x− 4 → f ′(x)= 2x+3

3. y = x−

23 → dy

dx = − 23x−

53

5.

v(r)= 4

3πr3 → v′(r)= 4πr2

7. y = x

2 + 4x+3x

= x3

2 + 4x1

2 +3x−1

2

dydx =

32x

12 + 2x

−12 − 3

2x−3

2 = 3 x2

+ 2x− 3

2 x3

9. z = Ay10 + Be

y = Ay−10 + Bey

′z = −10Ay−11 + Bey 11. 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( )

13.   Find the equation of the tangent line to f (x) = x5 − 5x +1 at x = −2 .

f −2( ) = −21 ′f x( ) = 5x4 − 5⇒ ′f −2( ) = 75 y+ 21= 75 x+ 2( )

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15.   Find the equation of the tangent line to y = x2

2 + 4x at x =1 .

f 1( ) = 92 dydx = x+ 4⇒

dydx

⎤⎦⎥x=1

= 5

y− 92 = 5 x−1( )

  17.   Find the equations of the lines tangent and normal to f x( ) = 5− x3 + x at the point 1, 5( ) .

′f x( ) = −3x2 + 12 x−12 ⇒ ′f 1( ) = − 52

Tangent: y− 5 = − 52 x−1( )  

Normal: y− 5 = 25 x−1( )  

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1.2 Homework Solutions

1. ddx x3 + 4x−π( )−7⎡

⎣⎢⎤⎦⎥= −7 x3 + 4x−π( )−8 3x2 + 4( ) = −7 3x2 + 4( )

x3 + 4x−π( )8

3. f x( ) = 1+ 2x+ x34 ⇒ ′f x( ) = 14 1+ 2x+ x3( )−34 ⋅ 2+ 3x2( ) = 2+ 3x2

4 1+ 2x+ x3( )34

5. Given the graph below, find (a) u ' 1( ) if u = f g x( )( )

⇒ ′u x( ) = ′f g x( )( )⋅ ′g x( )⇒ ′u 1( ) = ′f g 1( )( )⋅ ′g 1( )⇒ ′u 1( ) = ′f 3( )⋅ −3( )⇒ ′u 1( ) = −25( )⋅ −3( ) = 65

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(b) v ' 1( ) if v = g f x( )( )

⇒ ′v x( ) = ′g f x( )( )⋅ ′f x( )⇒ ′v 1( ) = ′g f 1( )( )⋅ ′f 1( )⇒ ′v 1( ) = ′g 2( )⋅2⇒ dne

(c) w ' 1( ) if w = g g x( )( )

⇒ ′w x( ) = ′g g x( )( )⋅ ′g x( )⇒ ′w 1( ) = ′g g 1( )( )⋅ ′g 1( )⇒ ′w 1( ) = ′g 3( )⋅−3⇒ ′w 1( ) = 2 3⋅−3= −2

7. If g(2) = 3 and ′g (2) = −4 , find ′f (2) if f (x) = e g(x)( ) .

′f x( ) = eg x( ) ⋅ ′g x( )⇒ ′f 2( ) = eg 2( ) ⋅ ′g 2( )⇒ ′f 2( ) = e3 ⋅−4 = −4e3

9. !d !dx 3x2 − 4x+ 9⎡

⎣⎢⎤⎦⎥= ddx 3x2 − 4x+ 9( )12⎡

⎣⎢

⎤

⎦⎥

= 12 3x2 − 4x+ 9( )−12 ⋅ 6x− 4( ) = 3x− 2

3x2 − 4x+ 9

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1.3 Homework Solutions 1. x

2 + 4y2 =16 2x+ 8y dydx = 0⇒ 8y dydx = −2x⇒ dy

dx = −x4y

3. x+5y2 = y−16 1+10y dydx =

12 y−16( )−12 dydx

1= 12 y−16( )12

−10y⎛

⎝⎜⎜

⎞

⎠⎟⎟dydx

1=1− 20y y−16( )122 y−16( )12

⎛

⎝⎜⎜

⎞

⎠⎟⎟dydx

dydx =

2 y−16( )121− 20y y−16( )12

5. Find the equations of the lines tangent and normal to x2 − 2x+ y2 + 4y = 20 at the point 5,1( ) .

2x− 2+ 2y dydx + 4dydx = 0

2y+ 4( )dydx = 2− 2xdydx =

2− 2x2y+ 4 = 1− xy+ 2

mtan =dydx 5,1( )

= 1− 51+ 2 = −43

Tangent: y−1= − 43 x− 5( )  

Normal: y−1= 34 x− 5( )  

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1.4 Homework Solutions 1. sin 4y x=

′y = 4cos4x 3. y = a3 + cos3 x

′y = 3cos2 x ⋅ −sin x( ) = −3cos2 xsin x 5. f (t) = 1+ tant3

′f t( ) = 13 1+ tant( )−23 ⋅sec2 t = sec2 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( ) = 1

2+ sin x ⋅1ln10 ⋅cosx =

cosxln10 2+ sin x( )

13. Find the equation of the line tangent to y = x+ cosx  at the point

0,1( ) .

′f x( ) =1− sin x⇒ ′f 0( ) =1 y−1=1 x− 0( )  

 

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15. Find all points on the graph of y = 2sin x + sin2 x where the tangent line is horizontal.

′f x( ) = 2cosx+ 2sin xcosx = 0 cosx = 0

x = ±π2 ± 2πn⎧⎨⎩

or sin x = −1

x = 3π2 ± 2πn⎧

⎨⎩

Points where the tangent line is horizontal: π2 ± 2πn, 3

⎛⎝⎜

⎞⎠⎟, −π2 ± 2πn, 1⎛⎝⎜

⎞⎠⎟

17. Find the equation of the line tangent to y = 2

π x+ cos 4x( )   x =π2 .

dydx =

2π − 4sin 4x( )⇒m π

2⎛⎝⎜

⎞⎠⎟= 2π

y− 2 = 2π x− π2⎛⎝⎜

⎞⎠⎟

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1.5 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 x−1( )⋅secx tan x

sec2 x =sec2 x− tan x−1( )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 ⋅e−x2 ⋅ −2x( )+ e−x2 = e−x2 1− 2x2⎡⎣ ⎤⎦ 7. y = excosx ′y = excosx x ⋅−sin x+ cosx⎡⎣ ⎤⎦ = e

xcosx cosx− xsin x⎡⎣ ⎤⎦

9. y = xsin 1

x

′y = x ⋅cos 1x ⋅−1x2

+ sin 1x = − 1x cos1x + 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

13. Find the equation of the line tangent to y = x2e−x at the point

1, 1

e( ) .

′y = x2 ⋅e−x ⋅ −1( )+ e−x ⋅2x m = ′y x=1 = −1e +2e =

1e

y− 1e =1e x−1( )

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15. h(t) = 1+ x2

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+ t22( )16

1− 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 = ex sin 4x( ) + 2 when x = 0

y x=0 = e0sin0 + 2 = 3⇒ 0,3( )

′y = exsin4x x ⋅cos4x ⋅4 + sin4x⎡⎣ ⎤⎦

′y x=0 = e0 0 ⋅cos0 ⋅4 + sin0⎡⎣ ⎤⎦ = 0

y− 3= 0 x− 3( )⇒ y = 3