Math 21A
1 (40 pts.)
Brian Osserman
Practice Exam 2 Solutions
Differentiate the following functions. Show your work if you wish to receive
partial credit. You do not need to simplify your answers.
(a) f (x) =
√3x+7 .
x4 +1
Answer: Quotient rule:
√
3 x4 + 1 − (3x + 7)(1/2)(4x3 )(x4 + 1)−1/2
0
f (x) =
.
x4 + 1
Or, general product/power rule:
3x + 7
f (x) = √
x4 + 1
0
(b) f (x) =
3
1 4x3
−
3x + 7 2 x4 + 1
.
x2 +1
.
tan−1 (x2 +1)
Answer:
f 0 (x) =
(2x)(tan−1 (x2 + 1)) − (x2 + 1)(1/(1 + (x2 + 1)2 ))(2x)
.
(tan−1 (x2 + 1))2
√
(c) f (x) = e2x−1 x + 1.
Answer:
√
1
f 0 (x) = (2e2x−1 ) x + 1 + e2x−1 (x + 1)−1/2 .
2
(d) f (x) = sec2 (x + sin x).
Answer:
f 0 (x) = 2 sec(x + sin x) ·
d
sec(x + sin x)
dx
= 2 sec(x + sin x) · sec(x + sin x) tan(x + sin x)
= 2 sec2 (x + sin x) tan(x + sin x)(1 + cos x)
d
(x + sin x)
dx
(e) f (x) = (ln x)10 .
Answer:
f 0 (x) =
2 (15 pts.)
10(ln x)9
.
x
Suppose u(x) is a differentiable function of x, and let f (x) = xu(x) , for
x > 0. Find f 0 (x), in terms of x, u(x), and u0 (x). Hint: take the natural log
of both sides and use implicit differentiation.
Answer: Taking ln on both sides,
ln f (x) = ln xu(x) = u(x) ln x.
Taking
d
,
dx
f 0 (x)/f (x) = u0 (x) ln x + u(x)/x.
So
f 0 (x) = f (x)(u0 (x) ln x +
3 (15 pts.)
u(x)
u(x)
) = xu(x) (u0 (x) ln x +
).
x
x
Suppose f (x) = x3 − ax + b. Find a, b if (1, 1) is on the graph y = f (x),
and the tangent line to the graph at (1, 1) has slope −3.
Answer: If (1, 1) is on the graph, then we have f (1) = 1. This means
13 − a · 1 + b = 1, so a = b. Then f 0 (x) = 3x2 − a, so f 0 (1) = 3 − a. If
f 0 (1) = −3, then 3 − a = −3, so a = 6, and then also b = 6.
2
4 (15 pts.)
Find the equation of the line tangent to the curve defined implicitly by
pπ q1
−1
2
2 sin y = x at the point ( 2 , 2 ).
Answer: Use implicit differentiation:
2
dy
1
p
= 2x,
dx 1 − y 2
so
p
dy
= x 1 − y2,
dx
Plugging in for x and y, the tangent line has slope
r
√
πp
π
.
1 − 1/2 =
2
2
The equation for the tangent line is
r
r √ 1
π
π
y−
=
x−
.
2
2
2
5 (15 pts.)
Consider the function
f (x) =


x sin 1 : x 6= 0
x

0 :
.
x=0
Is f (x) differentiable at x = 0? Why or why not? If so, what is f 0 (0)?
Answer: No. x sin x1 hits both y = x and y = −x inside (−δ, δ) for any
δ > 0, so the slope of the secants hits both 1 and −1 for any δ. Or, calculate
from the definition:
f (0 + h) − f (0)
h→0
h
1
h sin h
= lim
h→0
h
1
= lim sin ,
h→0
h
f 0 (0) = lim
and we know this limit does not exist.
3