Calculus in a Week

8/20/2019 Calculus in a Week

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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

Calculus in a Week

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Transcript of Calculus in a Week