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Final Exam Problem Bank
Chapters 1-
Differentiation Rules to Remember!
- Derivative of a Constant: d dx (c) = 0
- Power Rule: d dx (xn) = nxn−^1
- Constant Multiple Rule: If c is a constant, then d dx [cf (x)] = cf ′(x)
- Sum Rule: d dx [f (x) + g(x)] = f ′(x) + g′(x)
- Product Rule: d dx [f^ (x)^ ·^ g(x)] =^ f^
′(x)g(x) + f (x)g′(x)
[
f (x) g(x)
]
f ′(x)g(x) − f (x)g′(x) g(x)g(x)
- Derivative of the Natural Exponential Function: d dx (ex) = ex
Derivatives of the six basic trigonometric functions.
d dx (sin^ x) = cos^ x d dx (csc x) = − csc x cot x
d dx (cos^ x) =^ −^ sin^ x d dx (sec x) = sec x tan x
d dx (tan^ x) = sec^
(^2) x
d dx (cot x) = − csc^2 x
The Chain Rule If f (u) is differentiable at u, and u = g(x) is differentiable at x, then the derivative of the composite function f ◦ g(x) = f (g(x)), with respect to x, is
d dx [f (g(x))] = f ′(g(x)) · g′(x) OR df dx = df du · du dx
Derivatives of Log and Exponential Functions If b > 0, then
- (^) dxd (logb x) = (^) x ln^1 b • (^) dxd (bx) = bx^ ln b.
Derivatives of the inverse trigonometric functions.
d dx (arcsin^ x) =^
√^1
1 − x^2
d dx (arccos^ x) =^ −^
√^1
1 − x^2
d dx (arctan^ x) =^
1 + x^2
- Evaluate the following derivatives; you need not simplify your answers.
(a) (^) dxd
[
x^2 log 3 (2x + 1)
]
Solution. We can use the change of base formula (see ??) to rewrite log 3 (2x + 1) = ln(2ln 3x+1) , so d dx
[
x^2 log 3 (2x + 1)
]
ln 3
d dx
x^2 · ln (2x + 1)
ln 3
x^2 · d dx
ln
2 x + 1
ln 3
x^2 · 1 2 x + 1 · 2 + 2x ln (2x + 1)
(b) If y = (tan 2)e
√cos t , find dydt. Solution. Since tan 2 is a constant,
dy dt = (tan 2) · d dt
(^) e
cos t
= (tan 2)e
√cos t · d dt
(cos t) 1 /^2
= (tan 2)e
√cos t · 1 2 (cos t)−^1 /^2 (− sin t)
(c) If g(x) = sec(x^2 ex), find g′(x). Solution. d dx
sec
x^2 ex^
= sec
x^2 ex
tan
x^2 ex
d dx
x^2 · ex^
= sec
x^2 ex
tan
x^2 ex
x^2 ex^ + ex^ · 2 x
(Remember we use green boxes to indicate that we’re using the Product Rule.)
(d) If f (t) = t 5 3
1 + et^
, find f ′(t). (Can you rewrite f so that you don’t need to use the Quotient Rule?) Solution. We can rewrite f (t) = 15 t(1 + et)−^1 /^3 , so
f ′(t) =
d dt
t ·
1 + et
=^1
t · d dt
1 + et
3
1 + et
t
1 + et
d dt
1 + et
1 + et
t
1 + et
et^ +
1 + et
Now differentiate both sides with respect to x, and using the chain and product rule:
1 f (x) · f ′(x) =
x · ln(ln x) + ln x ·
ln x
x f ′(x) = f (x) ·
x · ln(ln x) + ln x ·
ln x
x
f ′(x) = (ln x)ln^ x^ ·
x · ln(ln x) + ln x ·
ln x
x
(d) f (x) = 3
x(x−2) x^2 +. Solution. This is a problem where logarithmic differentiation is not necessary, but it is super helpful! We can use it to break up this complicated function in smaller pieces that are easier to take derivatives of.
ln[f (x)] = ln
x(x − 2) x^2 + 1
ln
x(x − 2) x^2 + 1
[
ln(x(x − 2)) − ln(x^2 + 1)
]
ln(x) + ln(x − 2) − ln(x^2 + 1)
Now we can differentiate with respect to x:
f ′(x) f (x) =
x +^
x − 2 −^
2 x x^2 + 1
f ′(x) = f (x) ·
[
x
x − 2 − 2 x x^2 + 1
)]
f ′(x) = 3
x(x − 2) x^2 + 1
[
x
x − 2
2 x x^2 + 1
)]
Note: if you use a different method for computing the derivative, it is possible that you get an answer that has a completely different form and it is non-trivial to verify that the two answers are actually the same. If you are interested, you can check out the answer Wolframalpha gives. That answer is actually the same as this one.
(e) u(x) = sin
tan−^1 (ln x)
Solution. We need to use the chain rule.
u(x) = sin
tan−^1 (ln x)
u′(x) = d dx
sin
tan−^1 (ln x)
= cos
tan−^1 (ln x)
d dx
tan−^1
ln x
= cos
tan−^1 (ln x)
1 + (ln x)^2
d dx (ln x)
= cos
tan−^1 (ln x)
1 + (ln x)^2 ·^
x
(f) a(t) = (t + 2) √^3 (t + 5)^7 t − 5 Solution. This function is the product and quotient of a bunch of functions, and so we should use logarithmic differentiation.
a(t) = (t^ + 2)
(^3) (t + 5) 7 √ t − 5
We take natural log on both sides:
ln(a(t)) = ln
(t + 2)^3 (t + 5)^7 √ t − 5
= ln((t + 2)^3 (t + 5)^7 ) − ln(
t − 5) = ln((t + 2)^3 ) + ln((t + 5)^7 ) − ln((t − 5)^1 /^2 ) = 3 ln (t + 2) + 7 ln (t + 5) −
ln (t − 5)
And now we can differentiate
a(t) · a′(t) = 3 t + 2
t + 5
2(t − 5) a′(t) = a(t) ·
t + 2
t + 5
2(t − 5)
(t + 2) √^3 (t + 5)^7 t − 5
t + 2 +^
t + 5 −^
2(t − 5)
(g) s(t) = log 2
log 3
et
Solve for g′(t):
g′(t) = (ln t)^3 t
[
ln t
]
So, f ′(t) = 3 t^ ln 3 + 3t^2 + (ln t)^3 t
[
ln t
]
(j) f (x) = log 3 (arctan(x) arcsin(x)). Solution. This needs the chain rule, and then the product rule for the derivative of the inside function:
f ′(x) =
arctan(x) arcsin(x)
ln(3)
[
1 + x^2 · arcsin(x) + arctan(x) ·
√^1
1 − x^2
]
ln(3) arctan(x)(1 + x^2 )
ln(3) arcsin(x)
1 − x^2
- Find the equation of the tangent line to the curve
e^2 x^ = sin(x^2 + 2y) + 1.
at the point (0,0). Solution. Let’s start by finding the slope of the tangent line. We know that the slope is just the value of (^) dxdy at the point (0, 0), so let’s use implicit differentiation to find this slope. We start with the equation for the curve:
e^2 x^ = sin(x^2 + 2y) + 1
Differentiate both sides with respect to x:
d dx
e 2 x
d dx
sin
x^2 + 2y
2 e^2 x^ = cos(x^2 + 2y) ·
2 x + 2 dy dx
Plug in x = 0, y = 0:
dy dx Solve for dydx : dy dx
So, the line has slope 1. Since (0, 0) is a point on the line, the equation of the line is y = x.
- Find the line tangent to the heart curve (x^2 + y^2 − 1)^3 − x^2 y^3 = 0 at (− 1 , 1).
− 1 1
− 1
1
Solution. We start with the equation defining the curve.
(x^2 + y^2 − 1)^3 − x^2 y^3 = 0
Differentiate both sides with respect to x:
d dx
x^2 + y^2 − 1
− x^2 · y^3
d dx
x^2 + y^2 − 1
d dx
x^2 + y 2 − 1
[
x^2 · d dx
y 3
]
3(x^2 + y^2 − 1)^2
2 x + 2y dy dx
− x^2 · 3 y^2 dy dx −^2 xy
Now, we can plug in x = −1, y = 1:
−2 + 2 dy dx
− 3 dy dx
3 dy dx
dy dx
Therefore, the slope of the tangent line is 43 , so the equation of the line is y − 1 =
(x + 1). (From the picture, we can see that the tangent line should indeed have positive slope.)
- Here are graphs of two functions, f (x) and g(x). If F (x) = f (g(x)), what is F ′(1)?
− 2 − 1 1 2
− 2
− 1
1
2
Solution. We are looking for the points on the curve where dydx = 0. Starting from the equation for the curve, we can implicitly differentiate with respect to x: d dx
x^3 + y^3
d dx
2 ·^ x^ ·^ y
3 x^2 + 3y^2 dy dx
=^9
y + x dy dx
We want points where dydx = 0:
3 x^2 + 3y^2 · 0 =^9 2 (y + x · 0)
3 x^2 =^9 2 y 2 3 x^2 = y
So, we need y = 23 x^2. However, we also need to have points that are actually on the curve, so we also want x^3 + y^3 = 92 xy. Plugging the former equation into the latter,
x^3 +
x^2
x
x^2
x^3 +
27 x
(^6) = 3x 3
8 27 x
(^6) − 2 x (^3) = 0
x^3
x^3 − 2
So, either x = 0 or x^3 = 274 , which happens when x = √ 334. Looking at the picture of the curve, we see that the point with x = 0 does not have a tangent line at all, but there is a point with a horizontal tangent line, which must be the point with x = √ 334. The corresponding y-value is (^23)
√ 33 4
= √ 332. So,
the only point on the curve with a horizontal tangent line is
- Suppose you want to compute derivatives of the following functions. For which would you want to use logarithmic differentiation? Which could you do without logarithmic differentiation? You do not need to actually compute these derivatives!
(a) (x^3 + 17 sin x)ln 3 Solution. We don’t need log differentiation here. This function is of the form [f (x)]b^ so to compute the derivative, we can use the chain rule! d dx
( x^3 + 17 sin x )ln 3
= ln 3(x^3 + 17 sin x)ln 3−^1 · (3x^2 + 17 cos x)
(b) 7
√x (^3) +
Solution. We don’t need log differentiation here. This function is of the form [b]f^ (x)^ so to compute the derivative, we can use the chain rule!
d dx
√x (^3) +
√x (^3) + ln 7 · d dx
x^3 + 14
√x (^3) + ln 7 ·
(x^3 + 14)−^1 /^2 · (3x^2 )
(c) (x^2 +
x)tan^ x Solution. Here logarithmic differentiation is essential because this function is of the form f (x)g(x). Let y = (x^2 +
x)tan^ x^ We will start by taking logs on both side, and then differ- entiate: ln y = ln
[
(x^2 +
x)tan^ x
]
ln y = tan x · ln(x^2 +
x)
Differentiating both sides with respect to x 1 y
dy dx = sec^2 x ln(x^2 +
x) + tan x ·
x^2 +
x
2 x +
x−^1 /^2
- Two cars are approaching an intersection. A red car, approaching from the north, is traveling 20 feet per second and is currently 60 feet from the intersection. A blue car, approaching from the west, is traveling 30 feet per second and is currently 80 feet from the intersection. At this moment, is the distance between the two cars increasing or decreasing? How quickly? Solution. Intuitively, since both cars are heading toward the intersection, they are getting closer to each other, so the distance between the two cars should be decreasing. At the moment described in the problem, the situation looks like this (the red and blue car are represented by dots; the intersection is the +):
60
80
20 ft/s
30 ft/s
Now, we differentiate the relating equation with respect to t, using implicit differentiation: d dt (x
(^2) + y (^2) ) = d dt (z
2 x dx dt
Since we care about what happens at a specific moment, we can plug in our “snapshot” information.
At the moment we are interested in, x = 80, y = 60, dx dt = −30, dy dt = −20, and z = 100 by the Pythagorean Theorem. Plugging this all in to the previous equation gives
2(80)(−30) + 2(60)(−20) = 2(100) dz dt ,
so dz dt = −36. In words, the distance between the cars is
decreasing at an instantaneous rate of 36 ft/s.
- An oil tank in the shape of an inverted cone has height 10 m and radius 6 m. When the oil is 5 m deep, the tank is leaking oil from the tip at a rate of 2 m^3 per day. How quickly is the height of the oil in the tank decreasing at this moment? Note: The volume of a cone of radius r and height h is 13 πr^2 h. Solution. As usual, let’s start by drawing two pictures, a generic picture that applies at any time, and a “snapshot” showing the situation at the time we care about. We’ll also label some variables in the generic picture (the things labeled in red are changing).
6 m
10 m
r
h
Generic
6 m
10 m 5 m
Snapshot
Note that the radius and height of the tank are constant, but the radius and height of the oil change as oil leaks out, so we assign variables to those. Next, we figure out what we know and what we want to know. What we know: If we let V represent the volume of oil (in cubic m) at time t (where t is measured in days), then dVdt = −2. (This rate is negative because the volume is decreasing.) What we want to know: dhdt
Now, we try to relate variables. In this case, since the rate we know is d dtV and the rate we want to know is d dth , we try to relate V to h. First, the volume V of oil is the volume of a cone with radius r and height h, so V = 13 πr^2 h. We’d like to get V in terms of just h, so we need to express r in terms of h. Here, the key is similar triangles:
6 m
r 10 m h
From the similar triangles shown in blue, we get
10 =^
r h , so^ r^ =
3 h
- Plugging this into our equation for V , we get
Equation relating our variables: V =
π
3 h 5
h = 3 π 25 h^3
Once we have our relating equation, we differentiate it. Since we are looking for dhdt , we should differentiate with respect to t. d dt (V ) = d dt
3 π 25 h^3
dV dt =^9 π 25 h^2 dh dt At the moment we’re interested in, dVdt = −2 and h = 5:
−2 = 9 π 25
dh dt
So, dh dt =^ −^
9 π. So, at the time we’re interested in, the height of oil is
decreasing at (^92) π m/day
- At noon, you are running to get to class and notice a friend 100 feet west of you, also running to class. If you are running south at a constant rate of 450 ft/min (approximately 5 mph) and your friend is running north at a constant rate of 350 ft/min (approximately 4 mph), how fast is the distance between you and your friend changing at 12:02 pm? Solution. As usual, let’s start by drawing two pictures, a generic picture that applies at any time, and a “snapshot” showing the situation at the time we care about. We’ll also label some variables in the generic picture. Between noon and 12:02, you run (450 ft/min)(2 min) = 900 ft, and your friend runs (350 ft/min)( min) = 700 ft.
- During a night run, an observer is standing 80 feet away from a long, straight fence when she notices a runner running along it, getting closer to her. She points her flashlight at him and keeps it on him as he runs. When the distance between her and the runner is 100 feet, he is running at 9 feet per second. At this moment, at what rate is she turning the flashlight to keep him illuminated? Include units in your answer. Solution. As usual, let’s start by drawing two pictures, a generic picture that applies at any time, and a “snapshot” showing the situation at the time we care about. We’ll also label some variables in the generic picture.
Snapshot Generic
fence
(^80 )
observer
runner fence
80
x
observer
runner
θ
Next, we figure out what we know and what we want to know. What we know: dxdt = −9 (negative because the man is getting closer to the observer, so the distance x is decreasing) What we want to know: dθdt
Now, we try to relate variables. In this case, since the rate we know is dxdt and the rate we want to know is dθdt , we try to relate x and θ. Equation relating our variables: tan θ = 80 x Once we have our relating equation, we differentiate it. Since we are looking for dθdt , we should differ- entiate with respect to t. d dt (tan θ) = d dt
( (^) x 80
sec^2 θ dθ dt
dx dt Now, we plug in our snapshot information. In our snapshot, we can see that cos θ = 10080 = 45 , so sec θ = 54 :
25 16
dθ dt =^
dθ dt
So, the observer must turn the flashlight at
radians per second.
- As you’re riding up an elevator, you spot a duck on the ground, waddling straight towards the base of the elevator. The elevator is rising at a speed of 10 feet per second, and the duck is moving at 5 feet per second towards the base of the elevator. As you pass the eighth floor, 100 feet up from the level of the river, the duck is 200 feet away from the base of the elevator. At this instant, at what rate is the distance between you and the duck changing? Solution. As usual, let’s start by drawing two pictures, a generic picture that applies at any time, and a “snapshot” showing the situation at the time we care about. We’ll also label some variables in the generic picture.
Snapshot Generic
200
100
duck
you
x
z y
duck
you
Next, we figure out what we know and what we want to know. What we know: dxdt = −5 (negative because x is decreasing) and dydt = 10 (positive because y is increasing) What we want to know: dzdt
Now, we try to relate variables. In this case, since the rates we know about are dxdt and dydt , and the rate we want to know is dzdt , we try to relate x, y, and z. Equation relating our variables: x^2 + y^2 = z^2 Once we have our relating equation, we differentiate it. Since we are looking for dzdt , we should differ- entiate with respect to t. d dt
x^2 + y^2
= d dt
z^2
2 x dx dt
- 2y dy dt = 2z dz dt Let’s divide by 2 to simplify the equation a little:
x dx dt
- y dy dt = z dz dt Plug in the snapshot information:
200 (−5) + 100(10) =
2002 + 100^2
dz dt Solving, dzdt = 0 ft/s.
- Kelly is flying a kite; the kite is 100 ft above the ground and moving horizontally away from Kelly. At precisely 1 pm, Kelly has let out 300 ft of string, and the amount of string let out is increasing at a rate of 5 ft/s. If Kelly is standing still, at what rate is the angle between the string and the vertical increasing at 1 pm? (You may assume that the string is stretched taut so that it is a straight line.)
(c) At what rate is the angle between the ladder and the ground changing then?
Solution. Caveat: We really want you to follow the problem solving framework that has been set up in the previous problems as you work through these. This solution is a sketch, and you should work to fill in the details you would need in order to get full credit on this problem! a) Denote the base of the triangle by x, and the height of the triangle by y. We have x^2 + y^2 = 13^2. Taking derivative with respect to time yields 2x dxdt + 2y dydt = 0. At the moment we are interested in, we know that dxdt = 5, x = 12 and y =
132 − 122 = 5, so 2 · 12 · 5 + 2 · 5 dydt = 0. So dydt = −12 ft/sec.
b) Denote the area of the triangle by A. We have A = xy 2. So dAdt =
dxdt y+x dydt
Plugging in everything we know from the previous part, dAdt = 5 ·5+12 2 · (−12)= − 59 .5 ft^2 /sec. c) Denote the angle by θ. We will use the equation tan θ = yx. After applying (^) dtd , we have sec^2 (θ) dθdt = dydt x− dxdt y x^2. At the moment we are interested in, we have cos(θ) =^ 12
So^ dθ dt =^ 122 132 − 12 ∗ 12 − 5 ∗ 5 122 =^ − 169 132 = − 1 radsec. (Side note: The angle must be measured is radians for this computation to work! If we used degrees instead of radians, we would have to change our trig derivative rule to [tan θ]′^ = 180 π sec^2 (θ).)
(a) Use linear approximation to estimate
- 5 without using a calculator. Draw a sketch to explain what you are doing. Solution. We need to first pick a function to approximate; let’s use f (x) =
x. We want to approximate f (24.5), so we should pick a value of a near 24.5 as a “base” for our approximation. Let’s use a = 25, since we can evaluate f (25) exactly. f ′(x) = 12 x−^1 /^2 , so f ′(a) = 101. Thus, our tangent line has slope 10. We also know that (25, 5) is a point on the tangent line, so the line has equation y − 5 = 101 (x − 25), or y = 5 + 101 (x − 25). Therefore,
x ≈ 5 + 101 (x − 25) for x near 25.
Plugging in x = 24.5 gives
Here’s a picture of our situation:
25 t
(b) From your sketch, you should be able to tell whether your approximation is an overestimate or underestimate. Which is it? Solution. From the sketch it is clear that our estimate is an overestimate.
(c) Explain your answer to (b) using a second derivative. Solution. As we’ve seen, it is the concavity of a function that decides whether linear approx-
imations give over-estimates or under-estimates. So far, we’ve been able to “see” for ourselves whether a function is concave up or down. But there is a more mathematical way to decide this. Recall our old table again: f f ′^ f ′′ Positive / Negative nothing — increasing / decreasing Positive / Negative nothing concave up / concave down increasing / decreasing Positive / Negative — concave up / concave down increasing / decreasing A function f is concave up exactly when the first derivative is increasing i.e. exactly when the second derivative is positive. So an easy way to check concavity, is simply to compute the second derivative. Here, f ′′(x) = − 41 x−^3 /^2 = − 41 · √^1 x 3. Since this is negative for any value of x, the function is concave down, and so the estimate is an over-estimate.
- Use linear approximation to estimate 94 /^3. Is your estimate too high or too low?
Solution. The idea of linear approximation is that the line tangent to a graph y = f (x) at x = a is a good approximation for the actual graph around x = a, so we start by picking a function f (x) and a value of x where we will find the tangent line. Since we are looking for 9^4 /^3 , it makes sense to pick as our function f (x) = x^4 /^3. Then, we want to know f (9). Let’s find the tangent line to f (x) at x = 8 because 8 is pretty close to 9, and we can evaluate f (8) exactly; it’s 8^4 /^3 = (8^1 /^3 )^4 = 2^4 = 16. Next, we find the tangent line. The slope of the tangent line will be given by f ′(8). Using the Power Rule, f ′(x) = 43 x^1 /^3 , so f ′(8) = 43 · 81 /^3 = 83. Thus, the tangent line has slope 83 ; we know (8, 16) is a point on the tangent line, so the tangent line has equation y − 16 = 83 (x − 8), or y = 16 + 83 (x − 8).
So, we know that x^4 /^3 ≈ 16 + 83 (x − 8) for x near 8. Plugging in x = 9 gives 9^4 /^3 ≈ 16 + 83 = 18 +
To figure out whether this estimate is too high or too low, let’s think about the concavity of f (x). The second derivative f ′′(x) is 49 x−^2 /^3 , which is positive on (0, ∞); therefore, f is concave up on (0, ∞). So, our tangent line approximation must have been under the graph of f , which means that our estimate was too low.
- In this problem, we’ll look at the cubic function f (x) = x^3 + 3x^2 + 1.
(a) Find all critical points of f.
Solution. f ′(x) = 3x^2 + 6x = 3(x^2 + 2x) = 3x(x + 2), so f ′(x) = 0 at x = − 2 , 0. (There are no points where f ′^ is undefined.)
(b) Make a sign chart for f ′, and use this to decide whether each of the critical points you found is a local minimum, a local maximum, or neither. Solution.
sign of f ¢
f
So we know that at x = −2 there is a local max and at x = 0 there is a local minimum. The above is a strategy used to classify critical point. The strategy can be boiled down to the following:
