Here are the solutions to problems 1, 2 and 5 of the practice test that I didn’t have
time to go over today:
1. In each case, find the derivative y0 = dy/dx:
• y = sin−1 ( x + e x )
By the chain rule,
y0 = p
• y=
p
1
1 − ( x + e x )2
· ( x + e x )0 = p
1 + ex
1 − ( x + e x )2
.
ln(cos x )
Applying the chain rule twice, we obtain
1
− tan x
1
1
y0 = p
· (cos x )0 = p
· (ln(cos x ))0 = p
·
.
2 ln(cos x )
2 ln(cos x ) cos x
2 ln(cos x )
• y = x tan x
We use logarithmic differentiation:
ln y = tan x · ln x =⇒
1
y0
= sec2 x · ln x + tan x · .
y
x
It follows that
0
y =
tan x
sec x · ln x +
x
2
y=
tan x
sec x · ln x +
x
2
x tan x .
2. Verify that the function f ( x ) = ln x + 2x3 is strictly increasing, hence one-to-one
for 0 < x < +∞. If f −1 denotes the inverse of f , find the value of ( f −1 )0 (2).
First note that for x > 0,
f 0 (x) =
1
+ 6x2 > 0
x
which proves f is strictly increasing, hence one-to-one, hence invertible on
(0, +∞). To find the derivative of the inverse function f −1 at 2, note that f (1) = 2
so f −1 (2) = 1. Hence
( f −1 ) 0 (2 ) =
1
f 0 ( f −1 (2))
=
1
f 0 (1)
1
= .
7
5. Let R be the region in the plane
bounded by the curves y = e x and y =
e− x and the line x = 3.
(i) Sketch R and find its area.
The region R is shown on the right.
We have
y=e x
Z 3
(e x − e−x ) dx
0
h
i3
= e x + e− x
area of R =
R
0
3
= e +e
−3
x=3
y=e -x
− 2.
(ii) Find the volume of the solid obtained by rotating R about the x-axis.
By the general formula (method of “washers”),
volume of this solid = π
=π
=π
=
Z 3
0
Z 3
0
[(e x )2 − (e−x )2 ] dx
(e2x − e−2x ) dx
e−2x i3
+
2
2 0
h e2x
π 6
( e + e −6 − 2 ).
2
(iii) Find the volume of the solid obtained by rotating R about the y-axis.
By the general formula (method of “cylindrical shells”),
volume of this solid = 2π
Z 3
0
x (e x − e− x ) dx.
To compute this integral, we can use integration by parts, either by breaking it
up into the integrals of xe x and xe− x , or treat them together as follows:
Z 3
h
i3 Z 3
x
−x
x
−x
x (e − e ) dx = |{z}
x (e + e ) −
1 (e x + e− x ) dx
|{z}
|{z}
| {z }
| {z } 0
| {z }
0
0
0
u
v0
u
u
v
h
= 3 ( e 3 + e −3 ) − e x − e − x
3
= 3( e + e
−3
3
) − (e − e
−3
v
i3
0
) = 2e3 + 4e−3 .
Thus, the volume of the solid is
2π (2e3 + 4e−3 ) = 4π (e3 + 2e−3 ).