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Quiz 10 for MATH 161 Calculus I at Millersville University, Quizzes of Calculus

This is a quiz from the department of mathematics at millersville university for the course math 161 calculus i, given on december 3, 2004. It contains questions on finding derivatives using properties of the natural logarithm and determining if a function has an inverse and finding its inverse if it exists.

Typology: Quizzes

Pre 2010

Uploaded on 08/18/2009

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Millersville University Name
Department of Mathematics
MATH 161 Calculus I , Quiz 10
December 3, 2004
Please answer the following questions. Your answers will be evaluated on their correctness,
completeness, and use of mathematical concepts we have covered. Please show all work and
write out your work neatly. Answers without supporting work will receive no credit.
1. Find the following derivative using properties of the natural logarithm as needed.
d
dx
ln sx3
x5+ 1
d
dx
ln sx3
x5+ 1
=d
dx "1
2ln x3
x5+ 1#
=1
2
d
dx "ln x3
x5+ 1#
=1
2
d
dx hln x3ln(x5+ 1)i
=1
2
d
dx h3 ln xln(x5+ 1)i
=3
2
d
dx [ln x]1
2
d
dx hln(x5+ 1)i
=3
2
1
x1
2
1
x5+ 1
d
dx hx5+ 1i
=3
2x1
2
1
x5+ 1 h5x4i
=3
2x5x4
2(x5+ 1)
pf2

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Millersville University Name Department of Mathematics MATH 161 Calculus I , Quiz 10 December 3, 2004

Please answer the following questions. Your answers will be evaluated on their correctness, completeness, and use of mathematical concepts we have covered. Please show all work and write out your work neatly. Answers without supporting work will receive no credit.

  1. Find the following derivative using properties of the natural logarithm as needed.

d dx

 ln

√ x^3 x^5 + 1

 

d dx

 ln

√ x^3 x^5 + 1

  (^) = d dx

[ 1 2

ln

x^3 x^5 + 1

]

d dx

[ ln

x^3 x^5 + 1

]

d dx

[ ln x^3 − ln(x^5 + 1)

]

d dx

[ 3 ln x − ln(x^5 + 1)

]

d dx

[ln x] −

d dx

[ ln(x^5 + 1)

]

x

x^5 + 1

d dx

[ x^5 + 1

]

2 x

x^5 + 1

[ 5 x^4

]

2 x

5 x^4 2(x^5 + 1)

  1. Determine whether the function given below has an inverse. If it does, find the inverse and sketch the function and its inverse on the axes given below.

f (x) =

x + 1

The domain of this function is x ≥ −1. Since

f ′(x) =

x + 1

0 for x > − 1

the function is one-to-one and hence invertible.

-1 0 1 2 3 4 x

0

1

2

3

4

y