Math 105: Review for Exam I - Solutions
Basic Derivatives
d
(constant)= 0
dx
d n
x = nxn−1
dx
d x
e = ex
dx
d x
b = (ln b)bx
dx
d
1
logb x =
dx
(ln b)x
Basic Antiderivatives
antiderivative of xn =
xn+1
+ C if n 6= −1
n+1
antiderivative of ex = ex + C
antiderivative of bx =
d
1
ln x =
dx
x
d
sin x = cos x
dx
d
cos x = − sin x
dx
d
[kf(x)] = kf 0 (x)
dx
d
[f(x) + g(x)] = f 0 (x) + g0 (x)
dx
antiderivative of
bx
+C
ln b
1
= ln |x| + C
x
antiderivative of sin x = − cos x + C
antiderivative of cos x = sin x + C
antiderivative of kf(x) = k · [antiderivative of f(x)]
antiderivative of [f(x) + g(x)] =antiderivative of f(x)+
antiderivative of g(x)
d
f(x)g(x) = f 0 (x)g(x) + f(x)g0 (x)
dx
d f(x)
f 0 (x)g(x) − f(x)g0 (x)
=
dx g(x)
[g(x)]2
Limit Definition of Derivative
f 0 (x) = lim
h→0
f(x + h) − f(x)
provided this limit exists
h
Graphical Relationships Between f, f 0 , and f 00
f
f0
f0
positive
X
X
Exponent Rules
• bxby = bx+y
• (bx )y = bxy
negative
X
X
increasing
positive
X
decreasing
negative
X
concave up
increasing
positive
Logarithm Rules
• logb (xy) = logb x + logb y
• logb (x/y) = logb x − logb y
x
•
b
= bx−y
by
• logb (xy ) = y logb x
concave down
decreasing
negative
1. Let f(x) =
x+1
x+1
. Note that f(x) =
.
2
x −1
(x + 1)(x − 1)
(a) What is the natural domain of f? (−∞, −1) ∪ (−1, 1) ∪ (1, ∞), i.e., all reals except x = ±1
(b) Where is f discontinuous? x = 1 and x = −1
(c) What is lim f(x)? 0.5
x→3
(d) What is lim f(x)? −0.5
x→−1
(e) What is lim f(x)? undefined
x→1
(f) What is lim f(x)? ∞
x→1+
2. Let f(x) = 3x2 − 2x.
(a) Compute the average rate of change of f on the interval [2,2.1].
f(2.1) − f(2)
9.03 − 8
=
= 10.3
2.1 − 2
0.1
(b) Using the limit definition of the derivative, find f 0 (x).
f(x + h) − f(x)
provided this limit exists
h→0
h
3(x + h)2 − 2(x + h) − (3x2 − 2x)
= lim
h→0
h
2
2
3x + 6xh + 3h − 2x − 2h − 3x2 + 2x
= lim
h→0
h
h(6x + 3h − 2)
= lim
h→0
h
= lim (6x + 3h − 2)
f 0 (x) = lim
h→0
= 6x − 2
(c) Find the equation of the tangent line to f at x = 2.
We want y = mx + b. m = f 0 (2) = 6 · 2 − 2 = 10, so y = 10x + b.
When x = 2, y = f(2) = 3 · 22 − 2 · 2 = 8.
Thus, 8 = 10 · 2 + b, so b = −12 and we have y = 10x − 12.
(d) How would the derivative of g(x) = f(x) + 5 compare to f 0 (x)?
The graph of y = f(x) + 5 is the graph of y = f(x) shifted vertically by 5 units, but this has no
effect on the slope of the graph, so g0 (x) = f 0 (x).
(e) How would the derivative of h(x) = 5f(x) compare to f 0 (x)?
The graph of y = 5f(x) is the graph of y = f(x) stretched vertically by a factor of 5; this also
results in slopes that are 5 times greater at any given x-value. Thus, h0(x) = 5f 0 (x).
d
Note that we get the same result by considering our derviative rule
[kf(x)] = kf 0 (x) where
dx
k = 5.
5. Suppose that T (t) gives the temperature in Lewiston as a function of time. In each of the following
situations, determine if the signs of T , T 0 , and T 00 are positive, negative, zero, or unknown.
(a) The
The
The
The
temperature
temperature
temperature
temperature
is
is
is
is
60 degrees and falling steadily.
60, so we know T is positive.
falling, so we know T 0 is negative.
falling steadily, so we know the graph is linear, and T 00 is zero.
(b) The temperature is rising more and more slowly.
We don’t know whether the temperature is above or below zero, so the sign of T is unknown.
The temperature is rising, so we know T 0 is positive.
The temperature is rising more and more slowly, so we know the graph of T is concave down, and
T 00 is negative.
(c) The temperature is −5 degrees and rising.
The temperature is −5, so we know T is negative.
The temperature is rising, so we know T 0 is positive.
We don’t know the concavity of the graph of T , so the sign of T ” is unknown.
6. An object has vertical velocity v(t) = t2 − 5t + 4 feet per second on the interval [0, 5]. A positive
velocity indicates the object is ascending, and a negative velocity indicates it is descending. At time
t = 0, the object is 50 feet above ground.
(a) When is the object ascending? When is it descending?
v(t) = (t − 1)(t − 4) = 0 when t = 1 and t = 4.
On the intervals [0, 1) and (4, 5], v0 is positive, so the object is ascending during these times.
On the interval (1, 4), v0 is negative, so the object is descending during this time.
(b) When is the object’s acceleration most positive?
The acceleration is v0 (t) = 2t − 5. This is an increasing function, so it is most positive at the right
endpoint of the interval, or at t = 5.
(c) What is the greatest height the object reaches?
From part (a), we know that the object is rising until t = 1 and again until t = 5, so the greatest
height must occur at one of those times.
t3 5t2
We can find the height by taking the antiderivative of v(t), which gives h(t) =
−
+ 4t + C.
3
2
Since we know h(0) = 50, we see that C = 50.
Now we compare the heights at t = 1 and t = 5.
h(1) = 311/6 and h(5) = 295/6, so the maximum height is 311/6 = 51.83̄ feet.
7. Find the derivatives of the following.
6
x
9
+ √
+ e8 + 10 + π11 + sin(12) + log13 x + 14 cos x
7
5
x
x
(a) y = 2 + 3x + x4 +
x
6
9
1
= x1 ; √
= 6x−1/7; 10 = 9x−10; 2, e8 , π11 , and sin(12) are constants.
7
5
5
x
x
1
1
−1 −8/7
+ 0 + 9(−10)x−11 + 0 + 0 +
= 0 + (ln 3)3x + 4x3 + + 6 ·
x
− 14 sin x
5
7
(ln 13)x
1
90
1
6
−
+
= (ln 3)3x + 4x3 + − √
− 14 sin x
5 7 7 x8 x11 (ln 13)x
Useful observations:
dy
dx
dy
dx
(b) y = x3 2x
dy
= 3x22x + (ln 2)2xx3
dx
(c) y =
ex − x
3x2 + ln 5
dy
(ex − 1)(3x2 + ln 5) − 6x(ex − x)
=
dx
(3x2 + ln 5)2
(d) y = ln(x5e6x ) = ln(x5 ) + ln(e6x ) = 5 ln x + 6x ln e = 5 ln x + 6x by log and exponent rules
dy
5
= +6
dx
x
(e) y = 23+4x = 2324x = 23(24 )x = 8(16x) by exponent rules
dy
= 8(ln 16)(16x)
dx
8. Find antiderivatives of the following.
(a) y = π + ex − 2(3x )
antiderivative is πx + ex −
(b) y = x4 − 5 sin x +
antiderivative is
2(3x )
+C
ln 3
6
x
x5
+ 5 cos x + 6 ln |x| + C
5
See old exams and quizzes at http://abacus.bates.edu/˜etowne/mathresources.html