S. F. Ellermeyer
MATH 1190 –Exam 3 (Version 2) Solutions
November 6, 2006
Name
Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what
you are doing (mathematically) and you must also express yourself clearly. In particular,
write answers to questions using correct notation and using complete sentences where
appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations,
written explanations of why you are doing these calculations, etc.). It is not su¢ cient to
just write down an “answer” with no explanation of how you arrived at that answer. As a
rule of thumb, the harder that I have to work to interpret what you are trying to say, the
less credit you will get. You may use your calculator on this exam but you may not use any
books or notes. This exam contains 7 problems but you only have to do six of them in order
to get full credit. You may choose any six to do and I will grade only those six (even if you
work on all 7). You must write "Do not grade" on the problem that you do not want me to
grade.
1. Let f (x) = cos (x). Prove that f 0 (x) =
sin (h)
= 1 and
h!0
h
lim
sin (x). You may use the facts that
cos (h)
h!0
h
lim
1
= 0.
(You do not have to prove the above two facts!)
Solution: For any …xed real number x,
f (x + h) f (x)
h!0
h
cos (x + h) cos (x)
= lim
h!0
h
cos (x) cos (h) sin (x) sin (h) cos (x)
= lim
h!0
h
cos (h) 1
sin (h)
= lim cos (x)
sin (x)
h!0
h
h
= cos (x) 0 sin (x) 1
= sin (x) .
f 0 (x) = lim
2. Use the Quotient Rule and any necessary trigonometric identities to show that
d
(sec (x)) = sec (x) tan (x) .
dx
(You may also use the facts that d=dx (sin (x)) = cos (x) and d=dx (cos (x)) =
1
sin (x).)
Solution:
d
d
1
(sec (x)) =
dx
dx cos (x)
d
d
cos (x) dx
(1) 1 dx
(cos (x))
=
cos2 (x)
cos (x) 0 1 ( sin (x))
=
cos2 (x)
sin (x)
1
=
cos (x) cos (x)
= sec (x) tan (x) .
3. Use the Chain Rule and any other necessary di¤erentiation rules to …nd the derivatives
of the following functions. Do not simplify your answers.
(a) z = cos (x3 )
(b) z = tan3 (x)
(c) f (x) = (x + arctan (x))10
Solution:
(a)
dz
=
dx
sin x3
3x2
(b)
dz
= 3 tan2 (x) sec2 (x) .
dx
(c)
f 0 (x) = 10 (x + arctan (x))9
1+
1
1 + x2
.
4. A baseball diamond is a square with side 90 feet. A batter hits the ball and runs
toward …rst base with a speed of 25 feet/second. At what rate is his distance from
second base decreasing at the moment that he is two thirds of the way to …rst base?
(Your solution must be detailed, including all necessary calculations and a concluding
sentence that answers the question being asked).
2
Solution:
Let x be the runner’s distance from …rst base and let z be his distance from second
base. We know that
dx
= 25 feet/second
dt
and we want to …nd
dz
.
dt x=30
Since
z 2 = x2 + 902
at all times t, then we have
2z
dz
dx
= 2x
dt
dt
3
which gives us
dz
=
dt
25x
.
z
When x = 30, we have
z 2 = 302 + 902
and hence
z=
Thus
dz
dt
p
302 + 902
94:87 feet.
25 (30)
7:91 feet per second.
302 + 902
x=30
We conclude that the runner’s distance from second base is decreasing at a rate of
about 7:91 feet per second at the moment when the runner is two thirds of the way to
…rst base.
=p
5. For y de…ned implicitly as a function of x by
x2 + 2xy
y 2 + x = 2,
use implicit di¤erentiation to …nd dy=dx.
Solution:
d
x2 + 2xy
dx
y2 + x =
d
(2)
dx
gives
2x + 2x
dy
+ 2y
dx
2y
dy
+1=0
dx
which gives
(2x
or
2y)
dy
=
dx
dy
=
dx
2x
2y
1
2x 2y 1
.
2x 2y
6. Use any necessary di¤erentiation rules to …nd the derivatives of the following functions.
(a) f (x) = e
3x
.
(b) f (x) = ln (3x).
(c) f (x) = x ln (x).
Solution:
(a)
f 0 (x) = e
3x
d
( 3x) =
dx
3e
3x
(b)
f 0 (x) =
1 d
3
1
(3x) =
= .
3x dx
3x
x
4
.
(c)
d
d
(ln (x)) + ln (x)
(x)
dx
dx
1
+ ln (x) 1
=x
x
= 1 + ln (x) .
f 0 (x) = x
7. Use logarithmic di¤erentiation (and any necessary di¤erentiation formulas) to show
that the derivative of the function
y = xx
is
dy
= (1 + ln (x)) xx .
dx
Solution: We have
ln (y) = ln (xx )
so
ln (y) = x ln (x) .
Thus
d
d
(ln (y)) =
(x ln (x))
dx
dx
which gives
1 dy
1
=x
+ ln (x)
y dx
x
or
or
or
1 dy
= 1 + ln (x)
y dx
dy
= (1 + ln (x)) y
dx
dy
= (1 + ln (x)) xx .
dx
5