MTH 151: Practice Problems for Test 2 part 1
Sections for the test: 3.3–3.7 and 3.9
(1) Find the derivative. Do NOT simplify the answer
(a) f(t) = (3t+ 1)5 cos(5t2 + t− 3)
(b) f(y) = sin2(7y)
(c) f(x) = 5ex+2
x
+ 2epi
(d) f(x) = ln
(
(x2 + 3)2(x3 − 1)4√
x2 − 1
)
(e) f(x) =
ln(x2 + 3)
1− ln(x− 1)
(f) f(x) =
ex√
1 + x2
(g) f(x) = tan(x) cos(βx)eαx
(h) f(x) =
√
x3 − x5/2ex
x
√
x
(i) f(x) =
√
cos(
√
x secx)
(j) f(x) =
1
x2ex2
(k) y = tan−1(sin−1(x2 + x+ 1))
(l) y = log3
(
x
x2 + 1
)
(m) y =
√
1 + (log2(3x))
2
(n) y = (3x + x3 + 33)−1
(o) f(x) = a2 + tan(3x) sec(3x)
(p) y =
3
√
sin(pix) + tan(x2)
x2 + 1
(q) y = xe5 sin
2 x
(r) y = ln (x2(x3 + 1)4)
(s) f(x) = b2 + tan(sec(3x))
(t) f(u) = cos
(
eu − ue−u
eu + e−u
)
(u) y = cos(
√
tan(sin(4x5)))
(v) y = arctan(7e9x cos(x2 − 3x+ 2))
1
2(w) y = 5cos(pix) ln |4− 5 tanx|
(x) f(x) =
√
x+
√
x+
√
x
(y) y =
√
e3x + 7 cos(5x2) tan(log3(x))
(z) f(x) = ee + ex + xe + xx
( aa) f(t) = arcsin(
√
sin(3t))
( bb) g(x) = (2rarx + n)p , where a, r, n, p are constants.
(2) Find
dy
dx
by implicit differentiation.
(a) y2(2x− y) = cos(xy)
(b) ln(xy) = sin 2x+ cos 2y
(c) ex
2y =
√
x+ y
(d) xy = yx
(e) arcsin(x− y) = 1 + xy2
(f) ey cos(x) = 1 + sin(xy).
(g) y =
ey
1 + sin(x)
.
(3) Find the equation of the tangent line to the curve of
(a) y =
sinx
1− cosx at
(pi
2
, 1
)
(b) x+ y2 − 2xy = sin y + 1 at (1, 0)
(c) exy + 11 = 4x+ 5y at (3, 0).
(d)
√
x+ y +
√
xy =
3
2
at
(
1
2
,
1
2
)
(e) cos(x− y) + sin y = √2 at the point
(pi
2
,
pi
4
)
.
(f) y sin(8x) = x cos(2y) at the point
(pi
2
,
pi
4
)
.
(4) Use logarithmic differentiation to find the derivative of the function:
(a) y = (cosx)1/x
(b) y = (sinx)5
cos x
(c) y = (tanx)
√
x
3(d) y =
(x+ 1)2(2x2 − 3)√
x2 + 1
(e) y = (lnx)cosx
(f) y = (
√
x )
2x
(g) y = (cos(2x))cos(x)e−x
(h) y =
(
1 +
1
x
)2x
(i) y = (ln(x))x
2
(5) Find the points on the ellipse x2 + 2y2 = 1 where the tangent line has slope 1.
(6) Find equations of the tangent line to the curve y =
x− 1
x+ 1
that are parallel to the line x−2y = 2.
(7) Given the values f(3) = 4, g(3) = 7, f ′(3) = −1, g′(3) = 2. Find (fg)′(3) and (g/f)′(3).
(8) Find an equation of the tangent line to the curve y =
cos(x)
1− cos(x) at the point (a, f(a)) if a =
pi
3
(9) Suppose f
(pi
4
)
= −1 and f ′
(pi
4
)
= 2. Let g(x) = f(x) cosx and h(x) =
sinx
f(x)
. Find g′
(pi
4
)
and h′
(pi
4
)
.
(10) If h(x) =
√
4− 3 cos(pix)f(x), where f(1) = 7 and f ′(1) = 4, find h′(1).
(11) Find y′′ if
√
x+
√
y = 1.
(12) If xy + 9ey = 9e, find the value of y′′ at the point where x = 0.
(13) Find the derivative of the function and simplify the answer:
y = arctan
(√
1− x
1 + x
)
.
(14) Suppose f is a differentiable function such that f(g(x)) = x and f ′(x) = 1 + [f(x)]2. Show
that g′(x) = 1/(1 + x2).
(15) If F (x) = f(xf(xf(x))), where f(1) = 2, f(2) = 3, f ′(1) = 4, f ′(2) = 5, f ′(3) = 6. Find F ′(1).
(16) Given a function f(x), you are told that f ′(x) =
1
2x+ x2
. If H(x) = f(sinx), find H ′(x).
(17) The position of a particle is given by s(t) = t3 − 6t2 + 9t. Find the following:
(a) The velocity at time t.
(b) The acceleration at time t.
(c) When is the particle speeding up? When is it slowing down?
(d) The total distance traveled by the particle in the time interval 0 ≤ t ≤ 5
4(18) Solve a Related Rates problem: Use in-class examples and homework problems to practice.
Answers for Practice Problems for Test 2 part 1
(1) (a) f ′(t) = 5(3t+ 1)4(3) cos(5t2 + t− 3)− (3t+ 1)5 sin(5t2 + t− 3)(10t+ 1)
(b) f ′(y) = 2 sin(7y) cos(7y)(7)
(c) f ′(x) = 5ex+2x(1 + 2x(ln 2)) + 0
(d) Simplify first: f(x) = 2 ln(x2 + 3) + 4 ln(x3 − 1)− 1
2
ln(x2 − 1) Then
f ′(x) =
2(2x)
x2 + 3
+
4(3x2)
x3 − 1 −
2x
2(x2 − 1)
(e) f ′(x) =
(1− ln(x− 1)) 2x
x2 + 3
− ln(x2 + 3)
( −1
x− 1
)
(1− ln(x− 1))2
(f) f ′(x) =
ex
√
1 + x2 − 1
2
ex(1 + x2)−1/2(2x)
1 + x2
(g) f ′(x) = sec2(x) cos(βx)eαx − β tan(x) sin(βx)eαx + α tan(x) cos(βx)eαx
(h) f ′(x) = −ex − xex (Hint: Simplify the function before differentiating.)
(i) f ′(x) =
1
2
√
cos(
√
x secx)
(− sin(√x secx)) 1
2
√
x secx
(secx+ x secx tanx)
(j) f(x) =
1
x2ex2
= x−2e−x2 Then f ′(x) = −2x−3e−x2 + x−2e−x2(−2x)
(k) y′ =
1
1 + (sin−1(x2 + x+ 1))2
1√
1− (x2 + x+ 1)2 (2x+ 1)
(l) y = log3
(
x
x2 + 1
)
= log3(x)− log3(x2 + 1). Then y′ =
1
x ln 3
− 2x
(x2 + 1) ln 3
(m) y′ =
1
2
√
1 + (log2(3x))
2
(2 log2(3x))
1
x ln 2
(n) y′ = −(3x + x3 + 33)−2(3x ln 3 + 3x2)
(o) f ′(x) = 3 sec3(3x) + 3 tan2(3x) sec(3x)
(p) y′ =
1
3
(
sin(pix) + tan(x2)
x2 + 1
)−2/3
(x2 + 1)(pi cos(pix) + 2x sec2(x2))− (sin(pix) + tan(x2))(2x)
(x2 + 1)2
(q) y′ = e5(sinx)2 + xe5(sinx)2(10 sinx)(cosx)
5(r) Simplify to get y = 2 ln(x) + 4 ln(x3 + 1). Then y′ =
2
x
+
12x2
x3 + 1
(s) f ′(x) = sec2(sec(3x)) sec(3x) tan(3x)(3)
(t) f(u) = cos
(
eu − ue−u
eu + e−u
)
f ′(u) = − sin
(
eu − ue−u
eu + e−u
)
(eu + e−u)(eu − e−u + ue−u)− (eu − ue−u)(eu − e−u)
(eu + e−u)2
(2) (a)
dy
dx
=
−2y2 − y sin(xy)
4xy − 3y2 + x sin(xy)
(b) Rewrite ln(xy) = sin 2x + cos 2y as ln(x) + ln(y) = sin 2x + cos 2y and differentiate both
sides with respect to x. Solve for y′ to get
y′ =
2 cos(2x)− 1
x
1
y
+ 2 sin(2y)
(c) y′ =
1
2
√
x+ y
− 2xyex2y
x2ex2y − 1
2
√
x+ y
=
1− 4xyex2y√x+ y
2x2ex2y
√
x+ y − 1
(d) y′ =
y(x ln y − y)
x(y lnx− x)
(e) y′ =
(f) y′ =
(g) y′ = − y cosx
(1− y)(1 + sin(x)) .
(3) (a) y = −x+ pi/2 + 1
(b) y = 1/3(x− 1)
(c) y = −2x+ 6.
(d) y = −x+ 1
(e) y = 1/2x− pi/4
(f) y = −2x+ 5pi
4
(4) (a) y′ = −(cosx)1/x
(
ln(cosx)
x2
+
sinx
x cosx
)
(b) ln y = 5cosx ln(sinx).
y′ = (sinx)5cos x
(
ln(sinx) ln(5)(− sinx) + cosx
sinx
)
5cosx
(c) y′ = (tanx)
√
x
(
ln(tanx)
2
√
x
+
√
x
sec2 x
tanx
)
(d)
dy
dx
=
(x+ 1)2(2x2 − 3)√
x2 + 1
(
2
x+ 1
+
4x
2x2 − 3 −
x
x2 + 1
)
6(e)
dy
dx
= (lnx)cosx
(
− sinx ln(ln(x)) + cosx
x ln(x)
)
(f)
dy
dx
= (
√
x)2x(ln(x) + 1)
(g)
dy
dx
= (cos(2x))cos(x)e−x[sinx ln(cos(2x))− 2 cosx tan(2x)− 1]
(5)
(
2√
6
,− 1√
6
)
,
(
− 2√
6
,
1√
6
)
(6) y = x/2− 1/2, y = x/2 + 7/2
(7) (fg)′(3) = 1, (g/f)′(3) = 15/16
(8) y − 1 = −2√3(x− pi/3)
(9) g′(pi/4) = 3
√
2/2, f ′(pi/4) = −3√2/2
(10) h′(1) = 6/5
(11) y′′ =
1
2x
√
x
(12) y′′ =
1
81e2
.
(13) y′ =
−1
2
√
1− x2 .
17 Speeding up (1, 3), (5,∞) Slowing Down (0,1), (3,5), Distance traveled 39 m
7Problem 12. If xy + 9ey = 9e, find the value of y′′ at the point where x = 0.
Solution: If x = 0 in xy + 9ey = 9e, then we get 0 + ey = e, so y = 1 and the point where x = 0
is (0, 1). Differentiating implicitly with respect to x gives us
d
dx
(xy + 9ey) =
d
dx
(9e)
or
(1)y + x
dy
dx
+ 9ey
dy
dx
= 0.
Substituting 0 for x, 1 for y, and using y′ for
dy
dx
gives us 1 + 0 + 9e1y′ = 0, which implies 9ey′ = −1,
so y′ = −1/(9e).
Differentiating the expression from above,
(1)y + x
dy
dx
+ 9ey
dy
dx
= 0,
which can be written as
y + xy′ + 9eyy′ = 0,
implicitly with respect to x gives us
y′ + (1)y′ + xy′′ + 9eyy′y′ + 9eyy′′ = 0.
Now substitute 0 for x, 1 for y, and −1/(9e) for y′ and simplify to get 9ey′′ = 1
9e
or y′′ =
1
81e2
.
MTH 151: Practice Problems for Test 2 part 2
Sections for the test: 3.10, 4.1–4.5
(1) Find critical numbers and the absolute maximum and absolute minimum values of the function on
the given interval
(a) f(t) = 3
√
t
(
1− t
4
)
on [−8, 4].
(b) f(t) = |3t− 4| on [−1, 2]
(c) f(x) = 1− 2 cos(x) on the interval [0, pi]
(d) g(x) = sin(x)− sin2(x) on [0, 2pi]
(e) f(t) = t
√
4− t2 on [−1, 2].
(f) f(x) =
x− 1
x2 − x+ 1 on [−1, 1]
(g) g(x) = x1/3 − x−2/3 on [−8,−1]
(h) f(x) = x− lnx on [1/2, 2]
(i) f(x) = x4/5(x− 4)2 on [−1, 5]
(j) f(x) = x2e−3x on [0, 2]
(k) f(x) = x−2 lnx on [0.01, 4]
(l) f(x) =
x
x2 + 1
on [0, 2]
(m) f(x) = xe−x2/8 on [−1, 4]
(n) f(t) = 2− |t+ 1| on [−2, 1]
(o) f(x) = (x2 − 1)3 on [−1, 2]
(p) f(x) = e−x − e−2x on [0, 1]
(q) f(x) = ln(x2 + x+ 1) on [−1, 1]
(r) f(x) = |x+ 1|+ |x− 3| on [−2, 5].
(s) f(x) =
x√
x− 4 on [6, 12].
(2) Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then
find all numbers c that satisfy the conclusion of Rolle’s Theorem.
(a) f(x) = cos(2x) on [pi/8, 7pi/8]
(b) f(x) =
√
x− x
3
on [0, 9]
(3) Verify that the function satisfies two hypotheses of the Mean Value Theorem on the given interval.
Then find all numbers c that satisfy the conclusion of the Mean Value Theorem.
(a) f(x) = e−x/2 on [0, 2].
(b) f(x) =
x
x+ 2
on [1, 4]
(4) Tell yourself a joke.
(5) Use the first derivative test to locate the x-values at which function f(x) = x2/3(x− 5) has the local
minimum and maximum values.
(6) Find local and absolute maximum and minimum values of the function on the given interval
1
2(a) f(x) = x
√
1− x on [−1, 1]
(b) f(x) = (x2 + 2x)3 on [−2, 1]
(7) For each function f , do the following:
• Determine the intervals where the function f is increasing or decreasing;
• Determine intervals where f is concave upward or concave downward;
• Find local maximum and minimum values of f ;
• Find the points of inflection of the graph of the function.
(a) f(x) =
x2 + 1
x2 − 4
(b) f(x) =
x4 + 1
x2
(c) f(x) = x4(x− 1)3
(d) f(x) = sinx+ cosx on [0, 2pi]
(e) f(x) = e2x + e−x
(f) f(x) = 3x2/3 − x
(g) f(x) = ln(x4 + 27)
(h) f(x) = x1/3(x+ 4)
(i) f(x) = e−x2/2
(8) For each function f , do the following:
• Find the domain of the function f ;
• Find the vertical and horizontal asymptotes;
• Find the critical numbers;
• Determine the intervals where the function f is increasing or decreasing;
• Find local maximum and minimum values of f ;
• Determine intervals where f is concave upward or concave downward;
• Find the points of inflection of the graph of the function.
• Carefully sketch the graph
(a) f(x) =
x2 − 4
x2 − 2x
(b) f(x) =
x2
x2 + 9
(c) f(x) =
x√
x2 − 1
(d) Given that function f satisfies the following conditions:
f(0) = 0, f ′(−2) = f ′(1) = f ′(9) = 0, lim
x→∞ f(x) = 0, limx→6
f(x) = −∞,
f ′(x) < 0 on (−∞,−2), (1, 6) and (9,∞), f ′(x) > 0 on (−2, 1) and (6, 9),
f ′′(x) > 0 on (−∞, 0) and (12,∞), f ′′(x) < 0 on (0, 6) and (6, 12).
(e) Given that function f satisfies the following conditions:
lim
x→−∞ f(x) = 0, limx→4
f(x) =∞, f ′(−1) = f ′(2) = f ′(5) = 0,
f ′(x) > 0 for x in (−∞,−1), (2, 4), f ′(x) < 0 for x in (−1, 2), (4, 5), (5,∞),
f ′′(−2) = f ′′(1) = f ′′(2) = f ′′(5) = 0,
f ′′(x) > 0 for x in (−∞,−2), (1, 2), (2, 4), (4, 5), f ′′(x) < 0 for x in (−2, 1), (5,∞).
3(f)
f(x) =
2x3
x3 − 1 , f
′(x) =
−6x2
(x3 − 1)2 , f
′′(x) =
12x(2x3 + 1)
(x3 − 1)3 ,
(9) Use linear approximation to estimate the value
(a)
1
3
√
1002
.
(b) ln(0.99)
(c) e−0.015
(10) Evaluate the following limits. (Also look over problems we’ve done on webassign.)
(a) lim
x→0
sin(5x)
2x
(b) lim
x→0
arcsinx
x
(c) lim
x→0
ex − e−x − 2x
x− sinx
(d) lim
x→∞ x sin(pi/x)
(e) lim
x→∞
√
xe−x/2
(f) lim
x→0
tan(pix)
ln(1 + x)
(g) lim
x→∞ x tan
(
1
x
)
(h) lim
x→(pi/2)−
(tanx)cosx
(i) lim
x→0+
(1 + sin(3x))1/x
(j) lim
x→0
(
1
x
− 1
sinx
)
(k) lim
x→1
(2− x)tan(pix/2)
(l) lim
x→∞
(
2x− 3
2x+ 5
)2x+1

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