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Calculus in a Week1
Authored by
An anonyMOus source whose name is mentioned somewhere in the article
Note: Don’t expect too much out of this article2
Also, most of this was based off of Greg Kelly’s calc presentations (oops)
First off, this article assumes basic knowledge of trig. You should know what sinπ2
is. You should
also know sine and cosine sum identities if you want to derive the derivative of sine and cosine,but you don’t need to if you just want to take my word for it. Surprisingly, calculus rerquires littleknowledge of precalculus.
As usual, solutions are at the back.
Ok here we go
1Darn, a footnote in the title? Is that even legal?! Anyway, as the title implies, this is not a comprehensivecalculus course, and I have indeed skipped a good many of the sections. However, after you’re done reading this,you should have a pretty good grasp about the subject of calculus.
2Also note that although none of the info in here is false, this article is unreviewed and thus some of theteachings may be at least somewhat sketchy
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Contents
1 Brief intro to limits 5
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Squeeze Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Derivatives 7
2.1 Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Derivative Shortcuts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3.1 First Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3.3 Two more Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Derivative Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.5 Position, velocity, acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.6 Derivatives of Trig Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.7 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.7.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.8 Derivatives of Inverse Trig Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.9 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.9.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.10 Derivatives of Logs and Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.10.1 Return of the Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.10.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.11 Horizontal Tangents (maxima and minima) . . . . . . . . . . . . . . . . . . . . . . . 17
2.12 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.13 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.13.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3 Integration 22
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Contents AnonyMOus Calculus
3.1 First Steps of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.4 Reverse Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.4.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.5 u Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.6 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.6.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.7 Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.7.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.8 Volume (part I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.8.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.9 Disk/Washer Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.9.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.10 Cylindrical Shells Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.11 Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.11.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.12 L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.12.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.13 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4 Series 37
4.1 Some Notation and Early Series Stuff . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.2 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.2.1 nth Term Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.2.2 Possibilities for Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.2.3 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.2.4 Conditional Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.2.5 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.2.6 Alternating Series Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
4.2.7 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.3 p-series Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
4.4 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
4.4.1 One more incredible thing . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.4.2 Yep, that’s right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.5 Cool Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.5.1 e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.5.2 ln . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
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Calculus AnonyMOus Contents
4.5.3 tan inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5 Thanks 47
6 Solutions 48
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1
Brief intro to limits
1.1 Introduction
Ok
So all you really need to know about limits (for now) is basic substitution:Let’s look at
Example 1.1. limx→4 x2 + 3x − 4.
Solution. This is continuous (which roughly means that we can draw it without lifting our pencil),so we can just plug in 4.
limx→4
x2 + 3x − 4 = 16 + 12 − 4 = 24
Now, let’s look at this (usefulness! Dividing by 0 is possible?!):
Example 1.2.
limx→4
(x − 4)(x + 3)x − 4
Solution. Plugging in 4 gives us 00 !
That’s undefined! But this is a limit: x never EQUALS 4, it just approaches it. So we can simplify.
limx→4 x + 3 = 7
1.2 Squeeze Principle
Ok, this is pretty cool: So let’s say that
limx→a
f (x) = limx→a
g(x) = k
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Calculus AnonyMOus 1.2. Squeeze Principle
Also, let’s say that h(x) is between f (x) and g(x) in some interval containing a. Then, we havelimx→a h(x) = k as well!
How could such an obvious fact be useful?
Well, take this problem:
Example 1.3. Findlimx→0
x2 · sin
1
x
Solution. What do we do?
Well, we know that
−1 ≤ sin
1
x
≤ 1
Hey! We can multiply by x2! So, we have
−x2 ≤ x2 sin1x ≤ x2Taking the limit as x goes to 0, we get
limx→0
−x2 ≤ limx→0
x2 sin
1
x
≤ lim
x→0x2
Sincelimx→0
−x2 = limx→0
x2 = 0
we have
0 ≤ limx→0
x2 sin
1x
≤ 0
So,
limx→0
x2 sin
1
x
= 0
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2
Derivatives
This ”Part I” ”Part II” part might make it seem really long, but it isn’t that long (it reads prettyfast).
2.1 Part I
Just bear with me here; the beginning of this might be a little confusing. I’m going to ramble ona bit before teaching you any formulas.
I don’t think this is 100 percent accurate, but it’s certainly a good way to think of it:
Derivative = Slope
For instance, the derivative of the line 127894x − 783557849257820978860258 is 127894. So, thederivative of a line is it’s slope.
You might be thinking, ”slope is lines, you can’t find slopes anywhere else, how is derivative slope,i mean people take derivatives of curves and stuff this is IMPOSSIBLE YOU ARE WRONG”
Uh, no.
Let’s recall a definition. A tangent line/secant/curve/something is a line/secant/curve/somethingthat only touches something at one point (locally, which means it could touch the curve at anotherpoint, but not near the tangent point).
Look at that green line.
−4
−2
0
2
4
−4 −2 0 2 4
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Calculus AnonyMOus 2.2. Part II
The green line is a tangent line. How does this tie in to derivatives?
Well, let’s draw in a tangent line at all of the infinite points on the curve (not actually, but let’spretend that we do). Doesn’t it kind of represent the slope at that point?
Hmm
And the slopes of all of the points Represent the slope...
OF THE CURVE (or rather, we can find the slope at each point)!!!Summary:
Derivative = Slope
Curves have slopes, too!
Tangent = Intersects at 1 point
2.2 Part II
Back in Newton’s time, there were some controversies of who discovered what about calculus.Anyway, when you think of calculus, two names should leap to mind:
Liebniz (his first name is Gottfried; that’s cool)
Newton
Ok let’s get on topic again.
So Newton was doing some stuff
And he was like ”woah kewl !!1!111!one”
So we want to find the tangent line to the a curve
Let’s look at this:
−0.2 0.2 0.4 0.6 0.8 1 1.2 1.4−
0.2
0.4
0.6
0.8
1
1.2
f (x)
f (x + h)
f (x + h)
f (x + h)
f (x + h)A
B
C
D
E
Note that h is our variable here. x is a specific point. So, as h gets closer to 0, f (x + h) approaches
f (x). Since the points are (x, f (x)), and (x + h, f (x + h)), the slope is f (x+h)−f (x)x+h−x =
f (x+h)−f (x)h
We want h to approach 0, so let’s use a limit.
limh→0
f (x + h) − f (x)h
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Derivatives AnonyMOus Calculus
For a function f (x), the above is the derivative of f (x).
Let’s do an example or two.
Example 2.1. f (x) = x2
By our definition, the derivative is
limh→0
(x + h)2 − x2h
Simplifying,x2 + 2xh + h2 − x2
h =
2xh + h2
h =
h(2x + h)
h =
2x + h
1
Then, plugging in 0 for h, we find that our derivative is 2x (remember, h is APPROACHING 0,so we don’t have to worry about dividing by 0).
Use the formula limh→0f (x+h)−f (x)
h to answer THESE QUESTIONS!!!
2.2.1 Problems
Find the derivatives
Problem 2.1. x3
Problem 2.2. √
x
Problem 2.3.√
x3
Problem 2.4. x2 + 5x + 1
2.3 Derivative Shortcuts
Ok so there are are many ways to write derivativesdy
dx, ddx
, f (x), and others (replace x with whatever your variable is)
I’ll mostly use f (x) and dydx
. I’ll use f’(x) at first a lot, and then dydx
, and maybe ddx
a little.
Also, remember that the ”dx” means we’re differentiating with respect to x – if the function wasin terms of t or something, we would use dt.
So obviously those old guys were like ”Screw it this formula too much work”
So they were like ”swag formula making time”
2.3.1 First Formulas
All of these are easy to verify
This line is a bit confusing, but let’s say that α is a function in terms of x, and β is a function interms of x
For example, α could be 5x4, and β could be 67x5849 + 5478x4858249542894289 − 1 or something. Thepoint is, they’re both functions.
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Calculus AnonyMOus 2.3. Derivative Shortcuts
Then,
(α + β ) = (α) + (β )
Notice how we use α instead of f (α), since f (α) would mean the derivative of f (α)
Example 2.2. (5x4
+ 4x3
− 12x2
+ 3x + 1) = (5x4
) + (4x3
) − (12x2
) + (3x) + (1)
(c(α)) = c · (α)Note that c is a constant (like 6, 5, 23794, π, etc.)
Example 2.3. (5x4) = 5 · (x4)
Now: Remember that Derivative=Slope (I’ll always emphasize this, since it’s so dang important)
Therefore, (c) = 0 (where c is a constant)
Let’s end with a very important and influential rule that is very goodSuspense...
So funny story, actually, Leibniz had a good pal, who he named this rule after. It’s called:
The Obama Rule
...
Ok just kidding
It’s really called The Power Rule.
It states that:
(xn) = n · xn−1
This works for all real numbers n.Since this is so cool, here are the answers to the problems at the beginning, with very little effort.
Example 2.4. (x3) = 3x3−1 = 3x2
Example 2.5. (√
x) = (x12 ) = 1
2√ x
(yes, a lot of the time you will leave your denominator
unrationalized)
Example 2.6. (√ x3) = (x3
2 ) = 32√ x
Example 2.7. (x2 + 5x + 1) = (x2) + (5x) + (1) = 2x + 5
Wow now you’re like a pro in derivatives
(they grow up so fast)
But there is still much to learn
Now, some more problems.
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Derivatives AnonyMOus Calculus
2.3.2 Problems
Problem 2.5. x3 + 5x
Problem 2.6. 6x4 − 21x2 + x + 1x
Problem 2.7. 3√
x
Problem 2.8. 13√ x
Ok here we go
2.3.3 Two more Rules
The Product Rule
Ok, we have our functions α and β in terms of x again.
As you may have noticed, (αβ
) = (α
)(β
)So what is it?
I could use the derivative formula, but I’m lazy.
(αβ ) = (α)β + α(β )
For example, the derivative of (3x + 4)(2x + 1) is
(3x + 4)(2x + 1) + (3x + 4)(2x + 1) = 3(2x + 1) + 2(3x + 4) = 12x + 11.
Note that we COULD have expanded it out, but later in calculus, you’ll have to deal with functions
like x2
cos x or something, and you’ll have to use the product rule then.
The Quotient Rule
This one is a little harder (to use and remember)
It’s pretty hard to remember where the minus sign is, but here’s the quotient rule.
(αβ
) = β(α)−α(β)β2
Note that the bottom is β 2, and not the derivative of β 2
An easy way to remember which goes first is to take the derivative of 2x1
2.4 Derivative Orders
f (x) is the first derivative of f (x).
f (x) is the second derivative of f (x) (the derivative of the derivative).
f (x) is the third derivative of f (x) (the derivative of the derivative of the derivative).
After f , you usually start using numbers.
f (4)(x) is the fourth derivative of f(x), and f (n)(x) is the nth derivative of f (x).
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Calculus AnonyMOus 2.5. Position, velocity, acceleration
2.5 Position, velocity, acceleration
Position is position.
Velocity is the first derivative of position.
Acceleration is the second derivative of position
Let’s say the position of a moving particle is given by the equation y = x2
.The velocity of the particle is the derivative of x2, which is 2x.
The acceleration of the particle is 2.
Less important is the third derivative of position, jerk, and the fourth, fifth, and sixth derivativesthat are sometimes referred to as snap, crackle, and pop (no, I’m not kidding)
2.5.1 Problems
Don’t expand
Find the derivative:
Problem 2.9. (3x + 2)(2x + 1)
Problem 2.10. (8x2 + 9x + 4)(3x − 1)Problem 2.11. x
2−1x2−2x+1
Problem 2.12. x4−x2+x+1
x−4
2.6 Derivatives of Trig Functions
Ok this is real simple Recall that
sec x = 1
cos x
csc x = 1
sin x
cot x = 1
tan x——————————
(sin x) = cos x
(cos x) =−
sin x
(sec x) = sec x tan x
(csc x) = − csc x cot x(tan x) = sec2(x)
(cot x) = − csc2(x)A good way to remember this is ”All cats are evil” (credits to my math teacher)
(meaning: The derivative of all trig functions that start with c (and cats) have a negative sign infront of them)
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Derivatives AnonyMOus Calculus
2.7 Chain Rule
Let’s think We’re smart people yay we can find lots of derivatives
wait
whats the derivative of sin(x2)?
we don’t want to use the big fat limit formulawhat now?
No, not 2Chainz Rule
Ok let’s figure this out. So, we have
Example 2.8. y = sin(x2)
We want to find dydx
Let’s make u = x2 So dydu
= cos(u), and dudx
= 2x We want dydx
WAIT!
dy
du · du
dx =
dy
dx
So, our answer is cos(u) · 2x = cos(x2) · 2xLet’s try something else:
Example 2.9. Find the derivative of sin(cos(sec(cot(x))))
Hmmm let u = cos(sec(cot(x)))dy
du = cos(u)
du
dx =
...uh...Darn new variables Let’s say u2 = sec(cot(x))
du
du2= − sin(u2)du2
dx =
...uh...
Darn new variable Let’s say u3 = cot(x)
du2
du3= sec(u3)tan(u3)
du3
dx = − csc2(x)
Finally :/ Our answer is
cos(u) · − sin(u2) · sec(u3)tan(u3) · − csc2(x)This is
cos(cos(sec(cot(x)))) · sin(sec(cot(x))) · sec(cot(x)) · tan(cot(x)) · csc2(x)yay chain rule
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Calculus AnonyMOus2.8. Derivatives of Inverse Trig Functions
2.7.1 Problems
Here find the derivative go have fun
Problem 2.13. cos(sin(x))
Problem 2.14. sin(cos(x))
Problem 2.15. sin(√ x)Problem 2.16.
sin(x)
2.8 Derivatives of Inverse Trig Functions
Recall that sin−1(x), etc. doesn’t mean 1sinx , but rather the inverse sine (the angle for which sin(x)equals), like sin−1(1) = π
2, since sin
π2
= 1
So these are just some formulas When I say ”blahblahblah = blahblah” I mean the derivative of
blahblahblah is blahblah, not blahblahblah is equal to blahblah (oops just too lazy to write it)
sin−1(x) = 1√ 1−x2
cos−1(x) = − 1√ 1−x2
tan−1(x) = 11+x2cot−1(x) = − 1
1+x2
sec−1(x) = 1|x|√ x2−1csc−1(x) = − 1|x|√ x2−1
2.9 Implicit Differentiation
Darn. Were c00l. WAIT! CIRCLES?@!
What’s the slope of a circle? And ellipses and stuff but we like circles!!!!
Ok
Example 2.10. Let’s find the derivative of y2
So um the derivative of y is dydx
Errrrrrrrr let’s let u = y
Using the chain rule, we get the derivative is 2u · dydx
Since u = y , we get the derivative is 2y · dydx
So, to take the derivative of stuff with y, we just take it as if we were taking it as x, then tack ondy
dx at the end.
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Derivatives AnonyMOus Calculus
Example 2.11. Find the derivative of x2 + y2 = 1
We want to find the derivative, dydx
We take the derivative.
2x + 2y · dydx
= 0
(0 at the end since derivative of a constant is 0. Remember that, I always forget to make it 0 :P )
2y · dydx
= −2xDivide by 2y,
dy
dx = −x
y
So, just plug in the x and y coordinates of a given point, and we can find the slope at that point!
Using y and stuff is called implicit differentiation.
2.9.1 Problems
Find dydx
Problem 2.17. x2 + y2 = 4
Problem 2.18. x2 + y2 = 9 (notice anything weird about the answer to this and the answer to17?)
Problem 2.19. xy = 1
Problem 2.20. x + y = 1
Problem 2.21. x2 + y3 = 4
2.10 Derivatives of Logs and Exponentials
ex is pretty cool am i rite
Let’s use our formula one last time.
Example 2.12. Find the derivative of ex
limh→∞
ex+h − exh
= ex · eh − ex
h = ex ·
eh − 1
h
I’m too lazy to prove it, so I’ll just say that limh→0 eh−1h
= 1
OK
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Calculus AnonyMOus 2.10. Logs and Exponentials
So we have that the derivative of ex is
ex
Example 2.13. ax
ax = eln(ax) = ex·ln(a)
Then by the chain rule, the derivative is ax · dydx
(x · ln(a)). Since ln(a) is a constant, the derivativeof ax is ax ln(a)
REMEMBER:
It’s ax ln(a), not ax ln(x)
Example 2.14. Now let’s find the derivative of ln(x) using our handy-dandy implicit differ-entiation
y = ln(x)
ey = x
Taking the derivative,dy
dxey = 1
dy
dx = 1
ey
Ok, so that’s our derivative...or is it?! Remember that ey = x
So, ddx ln(x) =
1x
We can use similar methods to figure out that dydx
loga x = 1x ln(a)
REMEMBER:
The denominator is x ln(a), not x ln(x) !!!
2.10.1 Return of the Product Rule
Hey, we can also prove the product and quotient rules using implicit differentiation!
Product Rule: We want to find the derivative of
y = f (x) · g(x)Taking the natural log of this, and remembering that log ab = log a + log b, we get
ln(y) = ln(f (x)) + ln(g(x))
Taking the derivative, we get
dy
dx
y =
f (x)f (x)
+ g(x)
g(x) =
f (x) · g(x) + f (x) · g(x)f (x) · g(x)
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Derivatives AnonyMOus Calculus
Multiplying through by y, we get
dy
dx = f (x) · g(x) + f (x) · g(x)
as desired.
The proof of the quotient rule is similar.
2.10.2 Problems
Find the derivatives
Problem 2.22. 5x
Problem 2.23. x · ln xProblem 2.24. sin(ln x)
Problem 2.25. esinx
2.11 Horizontal Tangents (maxima and minima)
These are rather important, so they get a section for themselves!
Hm
Let’s say we are making chocolate chip pancakes
In fact, we own a chocolate chip pancake shop !
Our profit is described by −x2 + 12x−40, where x is the amount of pancakes we make (this makesno sense oops).
Obviously, our maximum is at x = 6, since that’s the vertex.But...what if the function was a cubic or something? Like −2x3 + 3x2 + 12x − 12 ?What would we do now? Well, you might want to use your graphing calculator and use the”maximum” button. But let’s imagine this is the 1800’s (Ye Olde Chocolate Chip Pancakes).
Let’s use...CALCULUS!!!
Ok, the maximum is the highest point in a function. That’s obvious.
If something is the highest point, it must rise on the way there, and fall on the way back. Lessobvious, but still pretty obvious.
If something is rising, its slope is positive. If something is falling, its slope is negative. Probably
not the first thing you would point out, but it makes sense.If slope is positive and then negative (or vice-versa), it must be 0 somewhere in-between. Yeah,that makes sense (this is actually a use of something called the Intermediate Value Theorem - afunction that is continuous that passes through two points must pass through all values in between.For example, if a continuous function passes through (1, 3) and (2, 5), there must be some pointon the function with y value 4. Or 4.2. Or 4.1829. Or any value between 3 and 5)
Now...here’s where you should say ”aha!”
If slope is neither rising nor falling, it must be a maximum or a minimum (finished going up, aboutto go down). But if something is neither rising nor falling, its slope must be 0!
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Calculus AnonyMOus 2.12. Related Rates
−4 −
3 −
2 −
1 1 2 3 4 5
−3
−2
−1
1
2
3
Since slope = derivative, the maximums and minimums of our function are the zeroesof it’s derivative!
Use the power rule, we take the derivative of our pancake function −2x3 + 3x2 + 12x− 12 and get
−6x + 6x + 12, which is equal to
−6(x + 1)(x
−2), so our maximum is at x = 2 (make sure to
check if it’s a maximum or a minimum! For instance, in our case, if f (2.1) > f (2) , it’s actually aminimum (luckily for us, it isn’t)!)
One more thing - It could be neither a maximum or a minimum. For instance, with y = x3, thefirst derivative is equal to 0 at 0 - but it’s neither a maximum nor a minimum (the slope goes up,to 0, and then up again)!
2.12 Related Rates
Ew. Practical applications.
Example 2.15. If a sphere has a radius of 20 cm, and its radius changes by .1 cm, approxi-mately how much does the volume change?
Let’s use calculus!
We have
V = 4
3π · r3
ThendV
dr = 4π · r2
(the change in volume with respect to the change in radius). We multiply by dr, and get
dV = 4π · r2drOk, we have a bunch of letters and numbers. But what can we do with them? What does thismean? WHAT DOES THIS MEAN???
Remember that dV is the change in volume, r is the radius, and dr is the change in radius. Hey,we have most of that! We know that the radius is 20, and the change in radius is .1, meaning
dV = 4π · r2dr
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Derivatives AnonyMOus Calculus
becomes
dV = 4π · 202 · .1 = 4 · 40 = 160π
So, the volume changes by 160π ≈ 502.65Wait, what? Using the volume formula for a sphere of radius 20.1 minus the volume of a sphereof radius 20, we get that the change is about 505, not 502...where did we go wrong?
Well, look at this (the curve here is x3, 43π · x3 is too steep at x = 20 lol)
−3 −2 −1 1 2 3 4 5 6 7−1
1
2
3
4
5
6
A
B
C
The curve is the curve. AB is the tangent line to the curve at 20 (just pretend it’s at 20, ok?:P ) Notice how we’re using the tangent line to approximate the function. So, instead of actuallycalculating the volume, we approximated it using the derivative (a.k.a. the slope!)
So it won’t be exact, but it’ll be close (as long as the change in radius is small. If we made thechange in radius 42 or something, it would be very inaccurate).
Now, different problem: Say the radius is 20 again, and it’s increasing at 1 cm/sec now (noticethat it’s a rate, not an amount now). How fast is the volume increasing?
The answer is 160π, because the instantaneous rate at which it is increasing is 160π cm3/sec whenthe radius is 10.
One more problem! This one’s sorta fun.
Example 2.16. Say we have a...errr...hot air baloon observatory (almost makes sense?) 500feet away from a spot right below a rising hot air balloon. Say that the angle between the hotair balloon, the observatory, and the spot below the balloon is π
4. If that angle changes by .1
radians per minute (dθdt
= .1), how fast is the balloon rising?
Solution. Look at this:
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Calculus AnonyMOus 2.13. Newton’s Method
500
h
thetaA B
C
We need an easy way to relate θ (wahhhh why didn’t it compile as θ D: ) and h. The easiest way,is, of course, tan θ = h500 ! Taking the derivative, we get
sec2 θdθ
dt =
1
500 · dh
dt
Why did we add the dθdt
and the dhdt
? Because, derivative is slope, and slope is rate of change, i.e.change over time! Hey, we know θ AND dθ
dt! They’re π
4 and .1, respectively. Plugging these in, we
have
sec2 π
4 · .1 = 1
500 · dh
dt
Since sec2 π4 = 2, we have
.2 = 1
500 · dh
dt
And thusdh
dt = 100
Superb!
2.13 Newton’s Method
Darn we can find slopes of functions and stuff
WAIT WHAT ABOUT AREAS UNDER FUNCTIONS
Wait lets save that for laterWAIT WHAT ABOUT APPROXIMATING ROOTS OF FUNCTIONS
Yeah we wanna do that
Let’s say we have a function like
x2 − 3Well
Like this
See this graph
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Derivatives AnonyMOus Calculus
AC
That function is x2 − 3Let’s call our A our initial guess – Make it something easy, like an integer
BC is the tangent line at (2, 1), which is y = 4x − 7HmNotice how AC is equal to f (2)
f (2)
Notice how C is an even better approximation than A.
So, given that our approximation is x, our newer, better approximation is x − f (x)f (x)
We plug in 2 to get 2 − f (2)f (2)
= 74We can repeat this over and over again to get even better guesses
So, for a function we want to approximate the roots of, we have that
gn+1 = gn−
f (gn)
f (gn)Where gn is the nth guess.
2.13.1 Problems
Problem 2.26. Estimate√
2 (hint: what is√
2 a root of?)
Now let’s get back to finding areas under curves.
What’s the area under (x − 1)2 from 0 to 1?
DarnWe could estimate it
Using rectangles
hm
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3
Integration
3.1 First Steps of Integration
Yay we’re finally at integration
Mainly because I skipped the majority of 2 chapters
Oh well no one likes those chapters
OK
so
If we draw 1 fat rectangle, that’s a pretty bad approximation of the area
If we draw 2 fat rectangles, it’s better but still bad
If we draw 10 skinny rectangles, it’s a good approximation
Wait...
What if we draw
INFINITE RECTANGLES
YES
So the height of each rectangle is f (x)
And the length of each is dx (the change in x, it’s infinitesimal and goes to 0)
Let’s introduce some notation
3.2 Notation
We have ba
f (x) dx
The a and the b represent the limits of integration – it means were finding the area from a to b.
The f (x) dx: f (x) is the height, dx is the width. So, we’re finding the area of a rectangle instandard ways.
Also, area under the x-axis is counted as negative area
Hmmmmm
We have the initial position
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Integration AnonyMOus Calculus
Then we uh
add up the slopes
because of instantaneous slope increases and stuff
and the the answer is
blah blah blah
long story shortAnti-differentiation is integration
First of all,
if f (x) is a function
F (x) is common notation representing the antiderivative
So anyway, ab
f (x) dx = F (a) − F (b)OK
Wait, one more thing.
3.3 C
The derivative of 2x2 is 4x.
The derivative of 2x2 + 1 is 4x.
The derivative of 2x2 + 10000 is 4x.
Uh-oh - They’re the same!
What do we do?
Usually, when we evaluate an antiderivative, we add +C, (C for constant), since a constant wouldbe eliminated after differentiation.
For example, the antiderivative of 4x is 2x2 + C
Let’s use the reverse power rule
3.4 Reverse Power Rule
The reverse power rule is exactly what it sounds like
The antiderivative of xa is xa+1
a+1
Except when a = −1. This is because the derivative of ln x = 1x , so the antiderivative of 1x isln x + C
Let’s solve a simple integral
Example 3.1. Find 53
x2 dx
This is
[13
x3 + C ]53
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Calculus AnonyMOus 3.4. Reverse Power Rule
Now, we plug in 5 and 3, and subtract.
Then1253
+ C − (273
+ C ) = 983
.
Hey, the +C s cancel out! I’m usually just going to not put in the +C when doing definiteintegration (definite integration is the one with numbers. For instance,
5
3 x2 dx = 98
3 (definite
integration), and x2 dx
=
1
3x3
+ C
(indefinite integration))The [13
x3]53 is an intermediate step. We just antidifferentiated the x2, and put the brackets and
numbers there to show we would plug them in later.
Sums and differences work too. For example,
Example 3.2. 53
x3 − x2 dx = 53
x3 dx − 53
x2 dx
Also,
Example 3.3. ab
f (x) dx = − ba
f (x) dx
Unsurprisingly,
Example 3.4.
ba
f (
x)
dx + cb
f (
x)
dx = ca
f (
x)
dx
There’s no one formula for finding the antiderivative of a function.
So have fun learning a few methods (you won’t)!
Also, remember to remember your derivatives
3.4.1 Problems
Problem 3.1. Find x3 + 5x2 dx
Problem 3.2. Find 51
3x2 + 2x − 1x2
dx
Problem 3.3. Find 5−3
8x3 + 6x2 − 1 dx
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Integration AnonyMOus Calculus
3.5 u Substitution
Ok
So first of all
b
a f (x) dx is a definite integral – we’re looking for a value
f (x) dx is an indefinite integralFor example, 53
x2 dx = 983 x2 dx = 1
3x3 + C
Now
Example 3.5. What’s
(x + 2)2 dx?
We make a clever substitution: u = x + 2 . We also know that du = dx
So, we have reduced it to
u2
du, which is u3
3 + C = (x+2)3
3 + C Let’s do another one
Example 3.6. Find
(2x + 1)2 dx
ok
try u = 2x + 1
we know that du = 2dx (since derivative is 1 to 2 ratio)
And 12du = dx
So, we have
u2 1
2 duThis is 12
u3
3 + C = u3
6 + C = (2x+1)3
6 + C
———————————————————————–
Sooooooooooooooooooooooooooo
Example 3.7. Find
sin x cos x dx
Hm.
What about u = sin x?
du = cos x · dxHey wait a minute
We have cos x dx in the problem!!!
So this becomes
u du
Which is simply u2
2 + C = sin2 x
2 + C
Ok
Long story short
Use u substitution if both the thing and its derivative are in the function
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Calculus AnonyMOus 3.6. Integration by Parts
3.5.1 Problems
Problem 3.4. Find
(sin x + x2)2 · (cos x + 2x) dxProblem 3.5. Find
esinx · cos x dx
Problem 3.6. Find √ 1 + 2xdxProblem 3.7. Find
1x·lnx dx
3.6 Integration by Parts
So uh
For functions u and v, we have
dy
dx
uv = u d
dx
v + v d
dx
u
That’s kinda hard to type. Let’s represent the derivatives of u and v be du and dv, respectively, just for convenience.
Integrating,
uv =
u dv +
v du
Or u dv = uv −
v du
We want u to be something we can easily take the derivative of, and dv to be something we can
easily integrateLet’s take an example
Example 3.8. Find
xex dx
We’ll let u = x, and dv = ex
Then, du = 1, and v = ex
So,
xex dx = xex − ex dxWhich is just xex
−ex
3.6.1 Problems
Problem 3.8. Find
sin x · ex dxProblem 3.9. Find
ln x dx
The end
Now we can get to stuff I actually like
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Integration AnonyMOus Calculus
OK
So we can find areas between stuff and the x axis
But what about areas between 2 curves?
EASY!!!
3.7 Areas
So it’s basically the same
Sum up infinite line segments
Example 3.9. What’s the area between y = 2 − x2 and y = −x?
Look at this:
−2 −1 1 2 3 4−1
1
2
3
A
B
so what’s the length of segment AB?
Let f (x) = 2 − x2, and g(x) = −xWe can see that the segment AB = f (x) − g(x) = (2 − x2) − (−x) = −x2 + x + 2We’re adding up the segments of length −x2 + x + 2We’re adding it up from −1 to 2So, our answer is
2−1−x2 + x + 2 dx = [−x
3
3 + x
2
2 + 2x]2−1 =
92
3.7.1 Problems
Problem 3.10. Find the area between y = x3 and y = 9x
Problem 3.11. Find the area between y = x2 and y = x3
Problem 3.12. Find the area in the second quadrant between y = 2 − x2 and y = −x
This section’s pretty interesting. So we’ve dealt with finding two-dimensional areas using infiniteline segments.
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Calculus AnonyMOus 3.8. Volume (part I)
Hm
What if we found three-dimensional areas using infinite two-dimensional shapes?
!!!
3.8 Volume (part I)
Example 3.10. Find the volume of a 3 by 3 by 3 cube
This is adding up infinite squares, right?
A
B
C
D
E
F
GH
I
J
K
L
We’re adding up infinite squares like JILK .
These squares each have area 9.
Let’s let A be at (0, 3), and H at (0, 0).
So, were adding up squares of area 9 from 0 to 3.
This is simply
30
9 dx = 27
Let’s try something different:
Example 3.11. Find the volume of a cone with base radius 5 and height 3.
Now we’re adding up infinite circles.
But these circles don’t have constant areas.
Their area is r2π, where r is the radius of the circle.
So how can we find r?
Well, at the vertex of the cone, the altitude is 3, and at the bottom, the altitude is 0 (as given inthe problem).
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Integration AnonyMOus Calculus
Also, r = 53
h
30
5
3h
2π dh =
25
9 π
30
h2 dh
(yes, you can take constants out and move them to the side)
This is 25
9 π · [ h
3
3 ]30 =
25
9 π · 9 = 25π
Note: This was not
30
r2π dr
We expressed the limits of integration in terms of the height, so we have to express the thing we’reintegrating in terms of it too.
Finally,
Example 3.12. Find the volume a square pyramid with height h and base side length k
We have that the proportion of side length to height is kh
The limits of integration are 0 and h.
Our cross-section is a square, and let’s call the distance from the vertex to the cross-section h (no,not the derivative of h. The side length of our cross-section, then, is k
h · h
So, our volume is
h
0
k
h ·h
2
dh = h
0
k2
h2
·(h)2 dh
k2
h2 is a constant, so we can move it to the side.
k2
h2
h0
(h)2 dh = k2
h2 · [ (h
)3
3 ]h0 =
k2
h2 · h
3
3 =
k2h
3
Which fits our definition of the volume of a pyramid: 13Bh
3.8.1 Problems
Problem 3.13. Find the volume of a sphere with radius r.
3.9 Disk/Washer Methods
So we can find the volume a regular solid
But what about say, the volume of when the area between the line y = x and the x-axis between0 and 5 is rotated about the x-axis?
OK
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Calculus AnonyMOus 3.9. Disk/Washer Methods
Let’s start small
What do we get when we rotate a vertical line segment that touches the x-axis around the x-axis?
A circle, of course!
And an area is simply adding up infinite line segments
So we’re simply adding up infinite circles!
Our limits of integration are 0 and 5.The radius of each circle is x (the height is x), and the area of each circle, then, is πx2
So, the volume is
50
πx2 dx
A lot of the time, you’ll just be asked to set up integrals and not evaluate them.
Now, let’s try the volume of sin x rotated about the x-axis from 0 to 2π
The radius of each circle is sin x, and the limits are 0 and 2π
So, our volume is 2π0
π(sin x)2 dx
So that’s the disk method, we’re adding up disks
Now for the washer method, which is somewhat similar.
Again, try a simple example.
Example 3.13. Find the area of when the line segment from (0, 1) to (0, 4) is rotated about
the x-axis
If it were from (0, 0) to (0, 4), it would simply be a circle of radius 4 – but we don’t have thesegment from (0, 0) to (0, 1) !
When we rotate from (0, 0) to (0, 4) we get a circle of radius 4, and when we rotate from (0 , 0) to(0, 1), we get a circle of radius 1.
So, the area is just 16π − π = 15πWe can add up infinite of these.
Example 3.14. Find the volume when the area between x2 and 2x is rotated about the x-axis.
We find when x2 = 2x. This is x = 0, 2. So, these are our limits of integration. In this range,2x > x2
So, the larger radius is 2x, and the smaller radius is x2
Therefore, the volume is 20
π(2x)2 − π(x2)2 dx
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Integration AnonyMOus Calculus
3.9.1 Problems
Problem 3.14. Find the volume when the area between x3 and the x-axis from 0 to 2 is rotatedaround the x-axis
Problem 3.15. Find the volume when the area between x and √
x is rotated around the x-axis.
3.10 Cylindrical Shells Method
OKAY so
The lateral surface area of a cylinder (the surface area minus the two bases) is 2 πr · h
Example 3.15. Find the volume of when the area between x2 and the x-axis from 0 to 2 isrotated about the y-axis
We could express the radii in terms of y and use the washer method, but let’s try something
different.Take a vertical line segment at a certain point, and rotate it around the y-axis. What do you get?The lateral surface area of a cylinder!
−3 −2 −1 1 2 3 4 5 6
−2−1
1
2
3
4
5
A
B
C
D
E
F
G H
I J
So, we’re adding up infinite lateral surface areas.
The radius of these cylinders is x, of course, and x2 is the height.
So, the volume is
2
0
2πx
·x2 dx
Darn, were great now
We have volume
Area
Slope
Root approximations
WAIT
What about...
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Calculus AnonyMOus 3.11. Length
3.11 Length
Take a look at this:
dx
dy
dx2 + dy2
A
B
C
So uh
AC = dx, the change in x.
BC = dy, the change in y.
...which means that AB =
dx2 + dy2
We’re adding up infinite ABs all on the length of our segment (lets say we’re finding the lengthfrom a to b)
Wait, adding up infinite stuff? That’s like...an INTEGRAL!!!
So, if we’re finding the length from a to b the answer is
ba
dx2 + dy2
Wait. Uh-oh. We don’t even know how to do that. Let’s try a little algebraic manipulation.Multiply everything by dx
dx
This becomes ba
dx2 + dy2 · dx
dx =
ba
dx2 + dy2 · dx√
dx2
Now that we have dx2 under a radical, we can move it into the big radical.
ba
dx2 + dy2dx2 dx =
ba
dx2dx2 +
dy2
dx2 dx
Finally, our formula is ba
1 +
dy
dx
2dx
Where, of course, dydx
is the derivative of our function.
Let’s look at an example.
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Integration AnonyMOus Calculus
Example 3.16. Find the length of x2 from 0 to 1.
Since dydx
= 2x, our answer is simply 10
√ 1 + 4x2 dx, which is...something. idk.
3.11.1 Problems
Problem 3.16. Find the length of 23
x32 from 2 to 4.
3.12 L’Hopital’s Rule
Darn. L’Hopital. L’Hospital. It’s all the same. Spelling is darn
There’s some old dude named Bernoulli (actually there was a whole squad of ’em, but his wasJohann Bernoulli)
Who was a pretty pro mathematicianAnd he came up with this pretty cool rule
Then he taught this guy named L’Hopital calculus
Then L’Hopital published the first calc book
And got credit
Anyway
OK
Let’s say f (x) = x2 − 4And g(x) = x
−2
And f (x)g(x) = x2−4x−2
Example 3.17. Let’s find limx→2f (x)
g(x)
Recall from the first calc thing that we can simplify it to limx→2 x + 2 = 4
2
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Calculus AnonyMOus 3.12. L’Hopital’s Rule
The steep function is f (x). The other one is g(x)
Notice that as x approaches 2, f (x)g(x)
becomes df dg
(change in f (x) over change in g(x)) =df
dxdg
dx
Wait
The limit of the ratio of the functions is the ratio of the derivatives=The ratio of their slopes?
YEAH
Let’s try this
limx→2
x2 − 4x − 2 =
2x
1 = 4
Wait, what about this?
limx→5 x, which is clearly 5. This is x1 . Take the derivatives, we get 10 = ∞ (well, ”equals infinity”
is wrong, but you get what I mean).
This is because L’Hopital’s only works for functions whose limit is indeterminate. Also, it onlyworks with fractions.
The indeterminate forms are
0
0, 0 · ∞, 00,∞0,∞−∞, ∞∞ , 1
∞
Example 3.18. Find
limx→∞
x sin 1
x
This approaches ∞ · 0
But how do we make it a fraction?Notice that x = 11
x
So, we have
limx→∞
x sin 1
x =
sin 1x
1x
Use L’Hopital’s Rule
limx→∞
cos 1x · − 1
x2
− 1
x2
Cancel out the − 1x2
,
limx→∞
cos 1
x = cos(0) = 1
One more example.
Example 3.19. Findlimx→0
(sin x)x
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Integration AnonyMOus Calculus
Darn, can we even make this a fraction?
YES WE CAN
Let our limit be L, we have elnL = L.
So, we can dolimx→0
(sin x)x = elimx→0 ln(sinx)x
Which, by our exponent rules, is
elimx→0 x·ln(sinx) = elimx→0
ln(sinx)1x
Using L’Hopitals,
elimx→0
1sinx ·cos
x
− 1x2 = elimx→0 −x
2· cosxsinx
Which is still in indeterminate form. Darn. Notice that cosxsinx
= 1tan x
So, we have
elimx→0 − x2
tanx = elimx→0 − 2x
sec2 x
Finally! We can plug in 0.
elimx→0 − 2x
sec2 x = e−01 = e0 = 1
3.12.1 Problems
Problem 3.17. Find limx→∞ 8x2+5x−3
3x2−42x+1337
Problem 3.18. Find limx→
01+2x3+5x
Problem 3.19. Find limx→0 xx
3.13 Improper Integrals
Darn so
We can find the area under 1x2
from 1 to 2 – that’s just 21
1x2
dx
But what about the area under 1x2
for x > 1? That’s infinite! We can’t do that...right?
Example 3.20. Find the area under 1x2
for x > 1
Well, let’s set up the integral I guess .-. ∞1
1
x2 dx = [−1
x]∞1
What now? We can’t just plug in infinity, that’s illegal...
HEY LET’S BE TRICKY
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Calculus AnonyMOus 3.13. Improper Integrals
We know limits, right???
Well...let’s replace ∞ by limb→∞ b ! YEAH!So we have
limb→∞
[−1x
]b1
We do our plugging in thingy
limb→∞
−1b − (−1
1) = lim
b→∞−1
b + 1
Clearly, limb→∞−1b = 0So, our answer is
0 + 1 = 1
One more example.
Example 3.21. Find the area under 1x
for x > 1
This is ∞1
1
x dx = [ln x]∞1
Using our little limit trick, this becomes
limb→∞
[ln x]b1 = limb→∞
ln b − ln1 = limb→∞
ln b − 0
Therefore, the area is infinite (it diverges).
...that’s about all there is to it
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4
Series
4.1 Some Notation and Early Series Stuff
OK SO
1 + 12 + 14 + ... converges (as the number of terms approaches infinity, the series approaches a certainvalue)
1 + 12
+ 13 + ... doesn’t (as the number of terms approaches infinity, the series DOESN’T approach
a certain value. This is why series like 1 − 1 + 1 − 1 + 1 − 1 + ... diverge)So in an infinite series
∞k=1
ak = a1 + a2 + a3 + a4 + ... + ak + ...
Partial sums:
S 1 = a1S 2 = a1 + a2
S n = a1 + a2 + ... + an
If S n converges as n approaches infinity, then it converges. If not, it diverges.∞n=1 ar
n−1 converges if −1 < r < 1, and the sum is a1−r (this is obvious, since it’s a geometricseries)
So all of this is pretty obvious
Now
Let a = 1 and r = x , such that
−1 < x
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Calculus AnonyMOus 4.2. Convergence
Wow
Let’s see this 11−xHm
HEY LET’S ANTIDIFFERENTIATE ALL THE TERMS
1 + x + x2 + x3 + ... dx =
11 − x dx
x + x2
2 +
x3
3 +
x4
4 + ... = ln(1 − x)
Wow
One more section before le super cool stuff
4.2 Convergence
So there are basically a ton of tests to see if stuff converges
So let’s say we have a series
a1 + a2 + a3 + a4 + ... =∞n=1
an
Heres one:
4.2.1 nth Term Test
So basically, if limn→∞ an = 0, the series diverges.This makes sense, right? For example, if our terms approach 1, then we’ll end up adding up infinite1s which is infinity.
That’s going to be a few series that you can rule out, but only a few...
4.2.2 Possibilities for Convergence
Let’s say our series is something like ∞n=0 x
n = 1 + x + x2 + x3 + x4 + ..., for some number x
There are a few possibilities:1. The series converges on a finite interval. This interval is the ”interval of convergence”
2. The series converges for all x
3. The series converges for some one number a
You’ll notice that there’s no ”diverges everywhere” – there has to be some number x so that itconverges.
Of course, we know 1 + x + x2 + x3 + x4 + ... only converges for −1 < x
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Series AnonyMOus Calculus
Direct Comparison Test
For a non-negative series, if every term of one series is less or equal to that of another series weknow converges, then the series converges.
For a non-negative series, if every term of one series is greater than or equal to that of anotherseries we know diverges, then the series diverges.
For example, we know that1
2 + 1 +
1
4 + 1 +
1
8 + 1 + ...
converges, since each term is less than 12n
4.2.3 Absolute Convergence
If
∞n=0 |an| converges, then
∞n=0 an converges. We says
∞n=0 an ”converges absolutely”.
Well, that makes sense.
4.2.4 Conditional Convergence
If the original sum, ∞
n=0 an, converges, but ∞
n=0 |an| doesn’t, then we say ∞
n=0 an ”convergesconditionally”
4.2.5 Ratio Test
We have our series∞n=0 an
Let’s set
L = limn→∞
an+1
an
If |L| 1, the series diverges.If L = 1, this test is inconclusive (darn)
Let’s do this problem:
Example 4.1. Find the values of x for which ∞
n=0(x − 6)n converges
We use the ratio test.
We find the values of n such that −1 < (x−6)n+1(x−6)n
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Calculus AnonyMOus 4.3. p-series Test
4.2.6 Alternating Series Test
If the terms of ∞
n=0 an decrease (i.e. an+1 < an for all n), then∞
n=0(−1)nan and∞
n=0(−1)n+1anconverge
This is just a way of saying that if the signs of the terms alternate, and the terms decrease, thenthe series converges
For example, 1 − 12 + 13 − 14 + 15 − 16 + ... converges.
4.2.7 Integral Test
So basically, let’s say we have our sequence, ∞
n=1 an, that is decreasing
Now let’s say each term can be described by a term in f (n)
For example, each term in 11
+ 14
+ 19
+ ... is 1n2
Now, if ∞1
f (n) converges, our series converges. If it diverges, our series diverges.
For example, we can determine if ∞n=1
1n3
converges or diverges by seeing if ∞1
1n3
dn converges.
Note that the bottom limit doesn’t have to be 1 – we can find if ∞n=5 1n3 converges by checking ∞
51n3
4.3 p-series Test
This will help with a few series.
∞
n=11
n p converges if p > 1
∞n=1
1
n p diverges if p = 1 or p
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Series AnonyMOus Calculus
OK
Now, if the first, second, third, and fourth derivatives of P (x) at 0 are the same as those of f (x)at 0, then our approximation is pretty dang good.
Math
f (x) = ln(1 + x)
f (0) = 0
P (0) = a0
Therefore,a0 = 0
f (x) = 1
1 + x
f (0) = 1
P (0) = a1
Therefore,a1 = 1
f (0) = − 1(1 + x)2
f (0) = −1P (0) = 2a2
Therefore,
a2 = −1
2
f (x) = 2 · 1(1 + x)3
f (0) = 2
P (0) = 6a3
Therefore,
a3 = 1
3
f (4)(x) = −6 · 1(1 + x)4f (4)(0) = −6
P (4)(0) = 24a4
Therefore,
a4 = −14
Thus,
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Calculus AnonyMOus 4.4. Taylor Series
P (x) = 0 + x − x2
2 +
x3
3 − x
4
4Look at this graph:
−1 1 2 3 4
−
−1
1
2
The approximation is pretty nice. Near 0, of course, which is what we were aiming for.
Long story short, we have Maclaurin Series, and Taylor Series.
Maclaurin Series are centered at 0.
Their general formula is
P (x) = f (0) + f (0)x + f (0)
2! x2 +
f (0)3!
x3 + ...
Taylor Series are centered at x = a, for some number a.
Their general formula is
P (x) = f (a) + f (a)(x − a) + f (a)2!
(x − a)2 + f (a)3!
(x − a)3 + ...
Example 4.3. Find the Maclurin Series for sin(x)
Using the formula,
f (0) = sin(0) = 0
f (0) = cos(0) = 1f (0) = − sin(0) = 0
f (0) = − cos(0) = −1The pattern just repeats from there.
So,
sin(x) = 0 + x + 0 − x3
3! + 0 +
x5
5! + 0 + ... = x − x
3
3! +
x5
5! − x
7
7! +
x9
9! − ...
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Series AnonyMOus Calculus
Heh, all the terms are of odd degree.
Heh, sin is an odd function (reflective over y-axis)
Here’s a graph of sin(x), then P (x) with terms being added on
−8 −6 −4 −2 2 4 6 8
−4
−2
2
Darn so
As terms get added on
It starts becoming a better approximation at more and more places
And then when we add infinite terms on
IT ACTUALLY BECOMES sin(x)
:o
FUNCTION = INFINITE POLYNOMIAL
GG
360NOSCOPEDarn
Ok, I’m not going to bore you with the math again for cos(x)
So here it is
cos(x) = 1 − x2
2! +
x4
4! − x
6
6! + ...
Which makes sense, since cos x is even.
We could also try summation notation.
sin(x) =
∞n=0
(
−1)n
·x2n+1
(2n + 1)!
cos(x) =∞n=0
(−1)n · x2n(2n)!
Also, unlike 11−x , 11+x
, ln(1+ x), etc., which only work from −1 < x
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Calculus AnonyMOus 4.4. Taylor Series
Example 4.4. Approximate cos(2x)
We can just plug in 2x for x!
cos(2x) = 1 − (2x)2
2! +
(2x)4
4! − ...
4.4.1 One more incredible thing
Example 4.5. Find the Maclurin Series for ex
Letting f (x) = ex
f (0) = e0 = 1
f (x) = ex
f (0) = e0 = 1 and so on.
So, the formula for ex is
1 + x + x2
2! + x3
3! + x4
4! + ... =∞
n=0xn
n!
Which is remarkably easy to memorize.
But that isn’t the cool part
Hi my name is Euler
I like math
Let’s do something which is technically not really allowed
eix = 1 + ix − x2
2! − ix
3
3! +
x4
4! +
ix5
5! − x
6
6! − ix
7
7! + ...
(we just plugged in ix for x in our ex formula)
Ew we have is and negative signs all over the place, this is nasty. Let’s rearrange some stuff (wecan do this because it converges – you’re not allowed to move around terms in infinite sequencesthat don’t converge).
eix = (1
− x2
2!
+ x4
4! − x6
6!
+ ...) + (ix
− ix3
3!
+ ix5
5! − ix7
7!
+ ...)
Factor out the is in the second part
eix = (1 − x2
2! +
x4
4! − x
6
6! + ...) + i(x − x
3
3! +
x5
5! − x
7
7! + ...)
Hmm...
Look familiar?
I’ll give you a hint – Look back up to the Taylor Series part.
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Series AnonyMOus Calculus
4.4.2 Yep, that’s right
1 − x22!
+ x4
4! − x6
6! + ... is our series for cosine, and x − x3
3! + x
5
5! − x7
7! + ... is our series for sine. In
other words, we have just showed that
eix = cos x + i sin x
Wait one more thing
So there are some pretty important numbers in math
And e,i,π, 1, 0 all probably rank in the top 10
Well let’s plug in π We geteiπ = cos π + i sin π = −1 + i · 0
So,
eiπ + 1 = 0
4.5 Cool Substitutions
These deserve a section for themselves :D
4.5.1 e
Using our series for e (and plugging in 1), we get the well known
e = 1 + 1
1! +
1
2! +
1
3! +
1
4! + ...
NICE
4.5.2 ln
Recall that 11+x
= 1 − x + x2 − x3 + x4 − x5 + ... Take the antiderivative,
ln(1 + x) = x − x2
2 +
x3
3 − x
4
4 +
x5
5 − x
6
6
Plug in x = 1
1 − 1
2 + 1
3 − 1
4 + 1
5 − 1
6 + ... = ln 2
W0w
4.5.3 tan inverse
We can use our handy-dandy Taylor Series formula to show that
tan−1(x) = x − x3
3 +
x5
5 − x
7
7 + ...
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Calculus AnonyMOus 4.6. Conclusion
(prove it for yourself!)
Going off on a tangent (hahaha), we note that this is remarkably similar to our series for sine, butwithout factorials in the denominator.
Anyway, we plug in 1, and we get
tan−1 1 = 1−
1
3 +
1
5 − 1
7 + ...
The number x for which tan x = 1 is π4
, which means tan−1 1 = π4
, so we get the nice sum
π
4 = 1 − 1
3 +
1
5 − 1
7 + ...
See if you can find more nice series!
4.6 Conclusion
Now you have a relatively good grasp of calculus, unless you just skipped the whole thing to readthis.
This isn’t a complete calculus course (otherwise it wouldn’t be called ”Calculus in a Week”), butit covers most of the stuff.
Feel free to ask questions
But I skipped a little
But I have to thank some people first so
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5
Thanks
Special thanks to:
• My Calc. teacher, who was actually very good
• Al• Mic• Ry• Rhy• Nat• Ad
• Al• Aks• ace• djmg• fridg• hsFTW
• and many others who helped put together this awful article.
Seriously though, with all the time you’ve wasted reading this, I thank you.
Sincerely,
mindless chores
OH YEAH SOLUTIONS
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6
Solutions
Solution. 2.1. We use the formula.
limh→0
(x + h)3 − x3h
= limh→0
x3 + 3hx2 + 3h2x − x3h
Simplifying and factoring,
limh→0
3h(x2 + hx)
h = lim
h→03(x2 + hx) = 3x2
Solution. 2.2. This one’s trickier.
limh→0
√ x + h −√ x
h
Now...we rationalize the numerator!!!
limh→0
√ x + h −√ x
h ·
√ x + h +
√ x√
x + h +√
x= lim
h→0x + h − x
h · (√ x + h + √ x) = limh→0h
h · (√ x + h + √ x)Simplifying,
limh→0
1√ x + h +
√ x
= 1
2√
x
Solution. 2.3. Similar to the last one,
limh→0
(x + h)3 −√ x3
h
limh→0
(x + h)3 −
√ x3
h ·
(x + h)3 +
√ x3
(x + h)3 +√
x3= lim
h→0(x + h)3 − x3
h ·
(x + h)3 +√
x3
Then,
limh→0
x3 + 3hx2 + 3h2x + h3 − x3h ·
(x + h)3 +√
x3 = lim
h→03h(x2 + xh)
h ·
(x + h)3 +√
x3
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Solutions AnonyMOus Calculus
Finally,
limh→0
3(x2 + xh) (x + h)3 +
√ x3
= 3x2
2√
x3
Since√
x3 = x32 ,
3x2
2 · x3
2
= 3
2 ·x
12 =
3
2 ·√
x
Solution. 2.4. We have
limh→0
(x + h)2 + 5(x + h) + 1 − x2 − 5x − 1h
= limh→0
x2 + 2xh + h + 5x + 5h + 1 − x2 − 5x − 1h
Which is
limh→0
2xh + 5h + h2
h = lim
h→0h(2x + 5 + h)
h = lim
h→02x + 5 + h = 2x + 5
Solution. 2.5. (x3 + 5x) = (x3) + (5x) Which, by the power rule is
3x3−1 + 5 · 1x1−1 = 3x2 + 5
Solution. 2.6. This is (6x4) − (21x2) + (x) + 1x
Which, by the power rule is
24x3
− 42x + 1 − 1
x2
Solution. 2.7 3√
x = x13 So, by the power rule, our answer is 13 · x−
23 (it’s okay to leave it like this –
or at least, my calculus teacher accepted a good many forms of answers)
Solution. 2.8 13√ x = x− 1
3 So, the power rule gives −13 · x− 43
Solution. 2.9 By the product rule, this is
(3x + 2)(2x + 1) + (3x + 2)(2x + 1) = 3(2x + 1) + 2(3x + 2) = 6x + 3 + 6x + 4 = 12x + 7
Which we can verify by expanding.
Solution. 2.10 Once again, we use the product rule.
(8x2+9x+4)(3x−1)+(8x2+9x+4)(3x−1) = (16x+9)(3x−1)+3(8x2+9x+4) = 72x2+38x+3
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Calculus AnonyMOus
Solution. 2.11 We use the quotient rule (darn I don’t really like it)
(x2 − 2x + 1)(x2 − 1) − (x2 − 1)(x2 − 2x + 1)(x2 − 2x + 1)2 =
(x2 − 2x + 1)(2x) − (x2 − 1)(2x − 2)(x2 − 2x + 1)2 =
−2x2 + 4x(x2 − 2x +
But apparently −2x2 + 4x − 2 = −2(x2 − 2x + 1), and x2 − 2x + 1 = (x − 1)2 So this is just−2
(x−1)2
Solution. 2.12 Use the quotient rule.
(x − 4)(x4 − x2 + x + 1) − (x − 4)(x4 − x2 + x + 1)(x − 4)2 =
(x − 4)(4x3 − 2x + 1) − (x4 − x2 + x + 1)(x − 4)2
Which is
4x4 − 2x2 + x − 16x3 + 8x − 4 − x4 + x2 − x − 1(x − 4)2 =
3x4 − 16x3 − x2 + 8x − 5(x − 4)2
Solution. 2.13 Use the chain rule, let u = sin(x) . dydu
= − sin u, and dudx
= cos x So, our answer is
− sin u · cos x
Plug u back in,
− sin(sin x) · cos x
Solution. 2.14 Similar to before, let u = cos x . dydu
= cos u, and dudx
= − sin x So, our answer is
cos u · − sin x = − cos(cos x) · sin x
Solution. 2.15 Let u =√
x . dydu
= cos u, and dudx
= 12√ x
So, our answer is
cos u
·
1
2√ x
= cos(√
x)
·
1
2√ x
Solution. 2.16 Let u = sin x . dydu
= 12√ u
and dudx
= cos x So, our answer is
1
2√
u · cos x = 1
2√
sin x· cos x
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Solutions AnonyMOus Calculus
Solution. 2.17 We have this as
2x + 2y · dydx
= 0
Then,
2y · dydx
= −2x
Dividing by y
on both sides, dy
dx = −x
y
Solution. 2.18 We have this as
2x + 2y · dydx
= 0
Then,
2y · dydx
= −2xDividing by y on both sides,
dy
dx = −x
y
DARN THE ANSWERS TO THIS ONE AND THE LAST ONE ARE THE SAME
Solution. 2.19 Using the product rule,
(x) · y + (y) · x = 0So,
y + dy
dx · x = 0Rearranging,
dy
dx = −y
x
Solution. 2.20 Take the derivative.
1 + dy
dx = 0
Sody
dx =
−1
Solution. 2.21 Take the derivative.
2x + dy
dx · 3y2 = 0
Sody
dx = − 2x
3y2
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Calculus AnonyMOus
Solution. 2.22 By the formula in this section, the derivative is
5x ln 5
Solution. 2.23 We use the product rule.
(x) ln x + (ln x)x = 1 · ln x + 1x · x = ln x + 1
Solution. 2.24 Using the chain rule, let u = ln x, so dydu
= cos u, and dudx
= 1x
So, our answer is
cos(ln x) · 1x
Solution. 2.25 Again, using the chain rule, let u = sin x, so dydu
= eu, and dudx
= cos x So, our answeris
esinx · cos x
Solution. 2.26 x2− 2 has a root of √ 2 !!! (x2) = 2x Our first guess is 1. f (1) = 2, and f (1) = −1Our next guess is, then, 1 − −12 (by the formula), which is 32f 32 = 3, and f 32 = 14 .Our next guess is, then, 32 − 112 ≈ 1.416 . We could go further, but let’s not.
Solution. 3.1 By the Reverse Power Rule, this is just
1
3 + 1 · x3+1 + 5
2 + 1 · x2+1 = 1
4x4 +
5
3x3
Solution. 3.2 We have
[x3 + x2 + 1x
]51 = 53 + 52 + 1
5 − (1 + 1 + 1) = 125 + 25 + 1
5 − 3 = 736
5
Solution. 3.3 We get
[2x4 + 2x3 − x]5−3 = 2 · 54 + 2 · 53 − 5 − (2 · (−3)4 + 2 · (−3)3 − (−3)) = 1384
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Solutions AnonyMOus Calculus
Solution. 3.4 We see that the derivative of sin x + x2 is in the problem. So, we make a substitution.Let u = sin x + x2, meaning du = (cos x + 2x) dx. So,
(sin x + x2)2 · (cos x + 2x) dx =
u2 du = u3
3 + C =
(sin x + x2)3
3 + C
Solution. 3.5 We note that the derivative of sin x is cos x, which is in the problem. So, we letu = sin x, meaning du = sin x dx. So,
esinx · cos x dx =
eu du = eu + C = esinx + C
Solution. 3.6 Let u = 1 + 2x. So, du = 2 dx or dx = 12
du. So, we have
√ 1 + 2x dx = √ u · 12
du = 23
u32 · 1
2 + C = 1
3u
32 + C = 1
3(1 + 2x)
32 + C
Solution. 3.7 We let u = ln x, meaning du = 1x
dx. So, 1
x ln x =
1
u du
Remembering that the antiderivative of 1u
is ln u, we have 1
u du = ln u + C = ln(ln x) + C
Solution. 3.8 Each are equally easy to differentiate and antidifferentiate, so let’s just let u = sin x,and dv = ex . So,
sin x · ex dx = sin x · ex −
cos x · ex dx
DARN, now we have another one :/ cos x ·ex dx. Let’s let u = cos x, and dv = ex. So,
cos x · ex dx = cos x · ex − − sin x · ex dx
Hey let’s do it again – WAIT!!! In the last part, we have − sin x · ex = − sin x · ex, which is our
original problem!
Let’s plug it in to our first equation sin x · ex dx = sin x · ex −
cos x · ex dx = sin x · ex − (cos x · ex −
− sin x · ex dx)
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Calculus AnonyMOus
Which is sin x · ex dx = sin x · ex − cos x · ex +
− sin x · ex dx = sin x · ex − cos x · ex −
sin x · ex dx
Add
sin x · ex dx to both sides.
2
sin x · ex dx = sin x · ex − cos x · ex
Which means sin x · ex dx = sin x · e
x − cos x · ex2
Solution. 3.9 This one’s tricky. The key step is noticing
ln x = 1 · ln x
Since we don’t know the antiderivative of ln x, we have no choice but to let u = ln x and dv = 1.So,
ln x dx =
1 · ln x dx = x · ln x −
x · 1
x dx = x · ln x −
1 = x ln x − x
Solution. 3.10 These intersect at 0 and 3, and 9x is greater in that range. So, the length of thesegment is 9x − x3, meaning that our area is
3
0
9x − x3 dx
Solution. 3.11 These intersect at 0 and 1, and x2 is greater in that range. So, the length of thesegment is x2 − x3, meaning our area is 1
0
x2 − x3 dx
Solution. 3.12 We have that the intersection points are
−1 and 2, and 2
−x2 is greater in that
range. However, the right limit of integration is 0, since the edge is the x-axis. So, our area is 0−1
(2 − x2) − (−x) dx
Solution. 3.13 Darn this is a little hard. So we have that the limits of integration are r and −r.However, how can we express the length of a cross section, which is a circle, in terms of r? Well,look at this.
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Solutions AnonyMOus Calculus
rh
√ r2 − h2
A
BC D E
So, we have that the radius of each cross-section is√
r2 − h2, meaning we can figure it out.The volume is
r
−rπ
·(√
r2
−h2)2 dh =
r
−rπ
·(r2
−h2) dh = π
r
−rr2
−h2 dh
Remember that r2 is a constant, so its antiderivative is r2h
π
r−r
r2 − h2 dh = π · [r2h − h3
3 ]r−r = π · (r3 −
r3
3 − ((−r)3 − (−r)
3
3 ) =
4
3πr3
Solution. 3.14 The radius of each circle is x3. So, our volume is
20
π(
x3
)
2 dx
Solution. 3.15 The intersections are at 0 and 1. Also, √
x > x in that range. So, our volume is 10
π(√
x)2 − π(x)2 dx
Solution. 3.16 We know dydx
=√
x. So, the length is
42
1 +
√ x2
dx =
42
√ 1 + x dx
. We can use u-substitution to get 42
√ 1 + x dx = [
2
3(1 + x)
32 ]42 =
2
3 · 5
√ 5 − 2
3 · 3
√ 3 =
10√
5
3 − 2
√ 3
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Calculus AnonyMOus
Solution. 3.17 This is ∞∞ . So, we use L’Hopitals
limx→∞
8x2 + 5x − 33x2 − 42x + 1337 = limx→∞
16x + 5
6x − 42Again, ∞∞ . So, we use L’Hopitals.
limx→∞
16x + 56x − 42 = limx→∞
166
= 83
Solution. 3.18 This is 13
. Hey, wait a minute! This isn’t in indeterminate form! So, our answer ismerely 13
Solution. 3.19 This is 00, so we can use L’Hopitals. We use the trick used earlier to make it afraction.
limx→
0xx = elimx→0 ln x
x
= elimx→0 x ln x = elimx→0 x ln x = elimx→0
lnx1x
Now we use L’Hopitals.
elimx→0
1x
− 1x2 = elimx→0 −x = e0 = 1
DONE
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