Quiz 1, Solutions
1. Q1 The function f (x) = 2x2 + ln x is one-to-one. Compute (f
Sol: We have
(f
1 0
(f
1 0
) (a) =
) (2).
1
f 0 (f 1 (a))
So when a = 2, we have
) (2) =
1 0
1
f 0 (f
1 (2))
To find f 1 (2), we set f 1 (2) = b. This is the same as saying that f (b) = 2 or 2b2 + ln b = 2 .
Now we have to guess a little. b = 1 is a likely candidate, we check it out; 2(1)2 + ln 1 = 2 + 0 = 2.
So f 1 (2) = 1.
Now we have
1
1
(f 1 )0 (2) = 0
=
f (f 1 (2))
f 0 (1)
We have f 0 (x) = 4x + x1 , so f 0 (1) = 4 + 1 = 5 and
(f
1 0
) (2) =
1. Q2 Di↵erentiate the function
f (x) =
1
=
f 0 (1)
1
5
(x 1)2 x3
.
(x2 + 1)2
Sol: We use logarithmic di↵erentiation. First we write out the equation y = f (x) and take the
logarithm of both sides.
(x 1)2 x3
y=
,
(x2 + 1)2
so
(x 1)2 x3
ln y = ln 2
= ln[(x 1)2 x3 ln(x2 + 1)2
(x + 1)2
= ln[(x
1)2 + ln x3
2 ln(x2 + 1)
= 2 ln(x
1) + 3 ln x
2 ln(x2 + 1)
Now we di↵erentiate both sides to get
1 dy
2
3
=
+
y dx
x 1 x
2
2x
.
+1
x2
Multiplying across by y, we get
h 2
dy
3
=y
+
dx
x 1 x
=
(x 1)2 x3 h 2
3
+
(x2 + 1)2 x 1 x
4x i
x2 + 1
4x i
.
x2 + 1