1
Practice exercises for Exam 2
Please remember that if there are homework problems or inclass examples that you have not understood you
would be better oﬀ revisiting them ﬁrst.
1. Implicit diﬀerentiation:
(a) Diﬀerentiate implicitly
e
xy
=
y
x
Answer:
e
xy
(
y
+
xy
0
) =
y
0
x

y
x
2
(b) Find the equation of the tangent line to the curve
√
y
=
√
x

2
y
+ 1 through the point where
y
= 1
Answer:
First plug in
y
= 1 to ﬁnd
x
at the points on the curve where
y
= 1. Then implicitly diﬀerentiate to
ﬁnd the slope. You’ll end up with a line with slope
1
10
through the point (4
,
1).
y
=
1
10
x
+
3
5
= 0
.
1
x
+ 0
.
6
2. A parametric curve is described by equations
x
(
t
) =
t
3
+ 1
y
(
t
) = ln(
t

1)
Find
y
and
dy
dx
when
x
= 9 in two diﬀerent ways:
(a) Directly
without determining the cartesian equation
for the function
y
(
x
).
Answer:
If you know the value of
x
then you can ﬁnd
t
. Recall
dy
dx
in terms of
dy
dt
and
dx
dt
. You’ll get 1
/
12
(b) By computing and diﬀerentiating the cartesian equation.
Hint: ﬁrst, solve for
t
from the equation for
x
, and plug
it into the expression for
y
.
Answer:
y
= ln(
3
√
x

1

1). Thus
y
0
=
1
3
√
x

1

1
·
1
3
(
x

1)

2
/
3
. Setting
x
= 9 again yields 1
/
12
3. Use linearization to approximate the following:
(a)
1
√
3
.
9
Answer:
The expression can be viewed as a function
f
(
x
) =
1
√
x
=
x

1
/
2
evaluated at 3
.
9. Note that
f
(4) is
much easier to ﬁnd, so instead of
f
(3
.
9) ﬁnd the tangent line to the graph of
f
(
x
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 Spring '08
 Plantz
 Derivative, Convex function, original function

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