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Math Test 2, October 29, 2008 - Differentiating Functions and Implicit Differentiation, Exams of Calculus

Virginia Polytechnic Institute and State University (Virginia Tech)Calculus

The answer key for math test 2 of class math 1205, held on october 29, 2008. It includes the differentiation of various functions using the power rule, quotient rule, and implicit differentiation. It also covers finding second derivatives and the equation of a normal line to an implicitly defined curve.

Typology: Exams

2019/2020

Uploaded on 11/25/2020

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Name ANSWER KEY Math 1 2 0 5 Tes t 2 Octo b e r 2 9 , 2 0 0 8
ID # Class Time or CRN
REA D T H E D I R E C T I O N S. YOU M U S T S H O W A L L W O R K O N THIS TEST A N D U S E M E T H O D S
LEA R N E D I N C L A S S T O RECEIVE F U L L C R E D I T . YOU MAY USE T H E C A L C U L A T O R S
DIS T R I B U T E D I N CLASS. (100 pt.)
1. Differentiate the following. Simplif y y o u r a n swers. (35 pt)
a.
′
y=20x3−1
x2+exe−1+ex+7xln 7 +2
3x
3
b. (use the quotient rule)
′
r(x)=(1+sin x)(1−sin x)−(x+cos x)(cos x)
(1+sin x)2=1−sin2x−xcos x−cos2x
(1+sin x)2=1−(sin2x+cos2x)−xcos x
(1+sin x)2
′
r(x)=1−1−xcos x
(1+sin x)2=−xcos x
(1+sin x)2
c.
y=ex3tan(8x)
4
′
y=ex3[1
4tan(8x)
( )
−3
4sec2(8 x)⋅8] +tan(8 x)
( )
1
4[3x2ex3]
′
y=2ex3tan(8x)
( )
−3
4sec2(8 x)+3x2ex3tan(8x)
( )
1
4
′
y=ex3tan(8x)
( )
−3
42sec2(8 x)+3x2tan(8 x)
( )
d.
y=ln x
x2+1
⎛
⎝
⎜⎞
⎠
⎟ ; x>0
y=ln x−ln(x2+1) =1
2ln x−ln(x2+1)
′
y=1
2x
−2x
x2+1
=1−3x2
2x(x2+1)
pf3
pf4

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Download Math Test 2, October 29, 2008 - Differentiating Functions and Implicit Differentiation and more Exams Calculus in PDF only on Docsity!

Name ANSWER KEY Math 1205 Test 2 October 29, 200 8

ID # Class Time or CRN

READ THE DIRECTIONS. YOU MUST SHOW ALL W ORK ON THIS TEST AND USE METHODS

LEARNED IN CLASS TO RECEIVE FULL CREDIT. YOU MAY USE THE CALCULATORS

DISTRIBUTED IN CLASS. (100 pt.)

  1. Differentiate the following. Sim plify your answers. (35 pt)

a.

y ′ = 20 x

3 −

x

2

  • ex

e − 1

  • e

x

  • 7

x ln 7 +

3 x

3

b. (use the quotient rule)

r ′ ( x ) =

( 1 + sin x )( 1 − sin x ) − ( x + cos x )(cos x )

( 1 + sin x )

2

1 − sin

2 x − x cos x − cos

2 x

( 1 + sin x )

2

1 − (sin

2 x + cos

2 x ) − x cos x

( 1 + sin x )

2

r ′ ( x ) =

1 − 1 − x cos x

( 1 + sin x )

2

− x cos x

( 1 + sin x )

2

c. y = e

x

3

tan( 8 x )

4

y ′ = e

x

3

[

1

4

( tan(^8 x ))

− 3 4 sec

2

( 8 x ) ⋅ 8 ] + ( tan( 8 x ))

1 4 [ 3 x

2 e

x

3

]

y ′ = 2 e

x

3

( tan( 8 x ))

− 3 4 sec

2 ( 8 x ) + 3 x

2 e

x

3

( tan( 8 x ))

1 4

y ′ = e

x

3

( tan( 8 x ))

− 3 4 2 sec

2 ( 8 x ) + 3 x

2

( tan(^8 x ))

d. y = ln

x

x

2

  • 1

; x > 0

y = ln x − ln( x

2

  • 1 ) =

ln x − ln( x

2

  • 1 )

y ′ =

2 x

2 x

x

2

  • 1

1 − 3 x

2

2 x ( x

2

  • 1 )
  1. Find the second derivative of (10pts)

h ′( x ) =

3 x

2

1 + ( x

3 )

2

3 x

2

1 + x

6

h ′′( x ) =

( 1 + x

6 )( 6 x ) − 3 x

2 ( 6 x

5 )

( 1 + x

6 )

2

6 x + 6 x

7 − 18 x

7

( 1 + x

6 )

2

6 x − 12 x

7

( 1 + x

6 )

2

3. The equation defines y as an implicit differentiable function of x and defines a

curve C in the plane (12pts)

a. Using implicit differentiation, find in terms of y and x.

2 x + 3 x

dy

dx

  • 3 y = 3 y

2

dy

dx

3 x

dy

dx

− 3 y

2 dy

dx

= − 2 x − 3 y

dy

dx

( 3 x − 3 y

2 ) = − 2 x − 3 y

dy

dx

− 2 x − 3 y

3 x − 3 y

2

2 x + 3 y

3 y

2 − 3 x

b. Write the equation of the line normal to C at (1,- 2 ).

m tan

⇒ m normal

⇒ normal line : y + 2 =

( x − 1 )

  1. Use logarithmic differentiation to differentiate. (11pts)

ln y = ln( x + cos x )

x

ln y = x ln( x + cos x )

y

dy

dx

= x

1 − sin x

x + cos x

  • ln( x + cos x )( 1 )

dy

dx

= y

x ( 1 − sin x )

x + cos x

  • ln( x + cos x )

dy

dx

= ( x + cos x )

x x ( 1 − sin x )

x + cos x

  • ln( x + cos x )
  1. Air is being let out of a spherical balloon at a rate of. When the diameter of the balloon

is 50 cm, how fast is the radius of the balloon changing? Use: (10pts)

V =

4 π

r

3

dV

dt

= 4 π r

2 dr

dt

− 100 = 4 π ( 25 )

2

dr

dt

2 ) π

dr

dt

dr

dt

25 π

cm

sec

cm

sec

The balloon’s radius is decreasing because

dr

dt

Pledge: I have neither given nor received help on this test.

Signed

dV

dt

cm

3

sec

r =

= 25 cm