Calculus I-01
Exam 3 Solutions
Write solutions completely for maximum credit. Put all work on separate
pages and keep this page for your reference. Each of the four problems is
worth 10 points.
1. Differentiate each of the following with respect to x:
a. xe−x
2
b.
ln(2x + 1)
2x + 1
−x2 )
d(xe
d e−x2 + e−x2 d(x ) , and since
=x
a. By the product rule,
dx
dx
dx
d(x ) = 1 and d e−x2 = e−x2 ( d (−x 2 ) = e−x2 (−2x ) , we get
dx
dx
dx
2
d(xe−x ) = xe−x2 (−2x ) + e−x2 = e−x2 (−2x + 1)
dx
2
b. By the quotient rule,
−1
2
1
) − ln(2x + 1)(2)( )(2x + 1) 2
d ln(2x + 1) =
2x + 1)
2
2x + 1
dx 2x + 1
−1
2
1
2x + 1(
) − ln(2x + 1)(2)( )(2x + 1) 2
2x + 1)
2
=
=
2x + 1
2
ln(2x + 1)
(
)−
2x + 1
2x + 1 = ( 2 − ln(2x + 1) )
2x + 1
(2x + 1) 2x + 1
2x + 1(
2. Integrate each of the following with respect to x:
a.
∫
cos( x 2)
dx , making the substitution u = 1+ sin( x )
2
1+ sin( x 2)
Let u = 1+ sin( x 2) . Then du = 12 cos( x 2)dx and 2du = cos( x 2)dx .
Substituting into the integral,
cos( x 2)
2
∫ 1+ sin(x ) dx = ∫ u du = 2ln | u | +c = 2ln | 1+ sin(x 2) | +c
2
To confirm the solution, differentiate the answer using the rule for
constant multiples and the chain rule twice to get:
d
d
(2ln | 1+ sin( x ) |) = 2 (ln | 1+ sin( x ) |) =
2
2
dx
dx
d
1
x ) d (x ) =
(1+ sin( x )) = 2
cos(
2
2 dx 2
1 + sin( x 2 ) dx
1 + sin( x 2 )
x
1
x )( 1 ) = cos( 2)
2
cos(
2 2 1 + sin( x )
1 + sin( x 2 )
2
2
1
1
∫ (2x +1)ln(2x +1) dx , making the substitution u = ln(2x +1)
Let u = ln(2x +1) . Then du = 1 (2)dx and 1 du = 1 dx .
b.
Substituting into the integral,
2x +1
2
2x +1
1
11
1
1
dx
=
du
= ln | u | +c = ln | ln(2x +1) | +c
∫ (2x +1)ln(2x +1) ∫ 2 u 2
2
To confirm the solution, differentiate the answer using the rule for
constant multiples and the chain rule three times to get:
d 1
( ln | ln(2x +1) |) =
dx 2
1d
1
1
d
(ln | ln(2x +1) |) =
(ln(2x +1)) =
2 dx
2 ln(2x + 1) dx
1
1
1 d
1
1
1
1
1
(2x +1) =
(2) =
2 ln(2x + 1) 2x + 1 dx
2 ln(2x + 1) 2x + 1
ln(2x + 1) 2x + 1
3. Suppose that f is a function defined by f (x) =
assume that f is 1-1.
4
,
1 + 2 ln x
and
a. Determine f '(x) .
Using the quotient rule, the rule for constants, and the rule for
sums we get:
f '(x) =
(1 + 2 ln x)
d
d
( 4) − 4 (1 + 2 ln x)
dx
dx
=
(1 + 2 ln x)
2
2
x
−4( )
(1 + 2 ln x)
2
=
−8
x(1 + 2 ln x)2
. Writing f(x)
as 4(1+ 2 ln x)−1 , and using the rule for constant multiples, the rule
for sums, and the chain rule also works well.
b. Determine a formula for f −1 (x)
Start by writing y =
4
1 + 2 ln x
and solve for x:
y(1 + 2 ln x) = 4 ⇒ y + 2y ln x = 4 ⇒ 2y ln x = 4 − y ⇒ ln x =
( 2y − 1 )
⇒x=e 2 ,
4
y 2 1
−
= −
2y 2y y 2
by using the definition of ln.
(2 − 1)
Then replace x by y and y by x to get: f −1 ( x) = e x 2
4. Assume that the equation xey + y =1 defines y implicitly as a
function of x.
a. Determine the value of
dy at the point (1,0).
dx
Differentiating each side of the equation and using various rules
of differentiation, we have:
d ( xey ) + dy = 0
dx
dx
dx
dx
dy
dy
dy
⇒ x d (ey ) + ey d ( x) + = 0 ⇒ xey + ey + = 0
dx
dx
dx
dx
dx
y
⇒ ( xey + 1) dy = −ey ⇒ dy = (xe−ey +1) . At the point (1,0), where x=1
dx
dx
0
dy = −e
−1
and y=0,
.
=
dx ((1)e0 +1) 2
xey + y =1⇒ d (xey + y) = d
(1) ⇒
b. Determine where the tangent line to the graph of the equation
xey + y =1 at the point (1,0) intersects the y-axis. (If you do not
get an answer to part a, state an assumption for the answer to
part a in order to answer this part of the problem.)
Since the tangent line passes through the point (1,0), if you go to
the left 1 unit from (1,0) to get to the y-axis, you would need then
to go up 1/2 unit to get to a point on the y-axis that is on the
tangent line, since the slope of the tangent line is –1/2.