MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH
SOLUTIONS
18.01 Calculus, Fall 2014
Professor: Jared Speck
Problem 1. Compute the second derivative of the function f (x) = arctan x.
x
Problem 2. Compute the derivative of the function f (x) = sin(x)xx .
Problem 3. Compute limx→π/2
L’Hôpital’s rule.
cos(3x)
.
cos(x)
Problem 4. Compute the derivative
dy
dx
At this point in the course, you are forbidden from using
for the curve x2 + y 3 + xy = 7 at the point (x, y) = (2, 1).
Problem 5. Find the equation for the tangent line to y = ln(x/3) at x = 3e.
Problem 6. Consider the function

sin x

− 2x
f (x) = 0

 cos x−1
x2
if x < 0,
if x = 0,
if x > 0.
Is f (x) continuous at x = 0? If not, then is the discontinuity removable?
Problem 7. Prove the quotient rule (u/v)0 = (u0 v −uv 0 )/v 2 using only the definition of a derivative.
Problem 8. Let r be a real number. Compute limh→0
derivative.
1
(1+2h)r −1
h
by interpreting this limit as a
2
MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS
Solutions
Problem 1. Compute the second derivative of the function f (x) = arctan x.
Solution:
d
1
arctan x =
,
(you should know how to prove this),
dx
1 + x2
d2
1
d
−2x
arctan
x
=
=
.
dx2
dx 1 + x2
(1 + x2 )2
x
Problem 2. Compute the derivative of the function f (x) = sin(x)xx .
Solution:
We first compute the derivative of xx using logarithmic differentiation:
y = xx ,
ln y = ln(xx ) = x ln x,
y0
= 1 + ln x,
y
y 0 = (1 + ln x)xx .
x
We now compute the derivative of xx :
x
z = xx ,
x
ln z = ln(xx ) = xx ln x,
d
d
1
z0
= ln x xx + xx ln x = ln x(1 + ln x)xx + xx ,
z
dx
dx
x
1
x
.
z 0 = xx ln x(1 + ln x)xx + xx
x
x
Finally, we compute the derivative of sin(x)xx using the product rule:
d
d x
x
x d
sin(x)xx = xx
sin(x) + sin(x) xx
dx
dx
dx
xx
= x cos x + sin(x)x
Problem 3. Compute limx→π/2
L’Hôpital’s rule.
Solution:
cos(3x)
.
cos(x)
xx
x
ln x(1 + ln x)x + x
x1
x
.
At this point in the course, you are forbidden from using
MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS
3
cos(3x) − cos(3π/2)
cos(3x)
= lim
x→π/2 cos(x) − cos(π/2)
x→π/2 cos(x)
cos(3x) − cos(3π/2)
x − π/2
= 3 lim
x→π/2
3x − 3π/2
cos(x) − cos(π/2)
cos(h) − cos(3π/2)
1
= 3 lim
×
cos(h)−cos(π/2)
h→3π/2
h − 3π/2
limh→π/2
lim
h−π/2
1
,
cos(u)|u=π/2
1
= 3(− sin(3π/2)) ×
− sin(π/2)
= 3(1)(−1) = −3.
=3
d
cos(u)|u=3π/2 ×
du
Problem 4. Compute the derivative
dy
dx
d
du
for the curve x2 + y 3 + xy = 7 at the point (x, y) = (2, 1).
Solution:
2x + 3y 2 y 0 + xy 0 + y = 0,
y0 = −
2x + y
5
= − = −1.
2
3y + x
5
Problem 5. Find the equation for the tangent line to y = f (x) = ln(x/3) at x = 3e.
Solution:
f (3e) = ln e = 1,
1
1
1
f 0 (x) =
× = ,
x/3 3
x
1
f 0 (3e) = .
3e
Tangent line (with x0 = 3e):
y − f (x0 ) = f 0 (x0 )(x − x0 ),
1
y − 1 = (x − 3e).
3e
Problem 6. Consider the function

sin x

− 2x
f (x) = 0

 cos x−1
x2
if x < 0,
if x = 0,
if x > 0.
4
MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS
Is f (x) continuous at x = 0? If not, then is the discontinuity removable?
Solution:
We recall the following important limits, which have been previously investigated in this course:
sin x
1
=− ,
x→0
2x
2
cos x − 1
1
lim+
=− .
2
x→0
x
2
Therefore, since limx→0 f (x) = − 12 6= f (0), f (x) is not continuous at x = 0.
However, if we redefine f by setting f (0) = − 12 , then for the new f, we have limx→0 f (x) = f (0).
Thus, the redefined f is continuous (and hence the original f has a removable discontinuity at
x = 0).
lim− −
Problem 7. Prove the quotient rule (u/v)0 = (u0 v −uv 0 )/v 2 using only the definition of a derivative.
Solution:
Given a point x and a small number ∆x, we define
∆u = u(x + ∆x) − u(x),
∆v = v(x + ∆x) − v(x).
Since u, v are differentiable (and therefore continuous), we have
lim ∆u = 0,
∆x→0
lim ∆v = 0,
∆x→0
∆u
= u0 ,
∆x→0 ∆x
∆v
= v0.
lim
∆x→0 ∆x
lim
Using the above facts, we have that
u 0
v
= lim
∆x→0
= lim
∆x→0
= lim
u+∆u
v+∆v
−
∆x
u
v
(u+∆u)v−(v+∆v)u
v(v+∆v)
∆x
v∆u−u∆v
v(v+∆v)
∆x
∆u
v
∆v
u
= lim
− lim
∆x→0 ∆x v(v + ∆v)
∆x→0 ∆x v(v + ∆v)
0
0
uv vu
= 2 − 2
v
v
u0 v − uv 0
=
.
v2
∆x→0
MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS
Problem 8. Let r be a real number. Compute limh→0
derivative.
(1+2h)r −1
h
by interpreting this limit as a
Solution:
(1 + 2h)r − 1
d
=
(1 + 2x)r |x=0 = r(1 + 2x)r−1 × 2|x=0 = 2r.
h→0
h
dx
lim
5