Question

Find the derivative of y(theta)= theta+cos(^2)(3theta)

Answer 1

so if you had an x instead of theta, do you think you could do it?

Answer 2

i got 1-3sin(3theta) but i want to know why the answer is 1-3sin(6theta)

Answer 3

chain rule use

Answer 4

ohk, well, I am not getting that answer because I donot know the double angle identities, but I can help ya through :)

Answer 5

so we have \[y(\theta)= \theta+cos^2(3\theta)\]

Answer 6

oh wow i dent know i had to use a double angle identity 0.o

Answer 7

dont need double angle

Answer 8

first thing I want to do is substitute @DanJS at the end, that is how they simplified it

Answer 9

(-6)*sin(3x)*cos(3x)

Answer 10

so @anarvizo, if we let \(u=3\theta\) can you tell me du=?

Answer 11

so i'd get 1-3sin(3 theta) still ;/ unless I'm missing dropping the ^2 in from of sin

Answer 12

that is the derivative of cos^2(3x)

Answer 13

okay lets see hang on

Answer 14

ic yeah you can simplify the answer with the double angle formula

Answer 15

(I am showing you how to get that derivative)

Answer 16

so what would du=?

Answer 17

du would just = 3 wouldn't it?

Answer 18

thats why i have the 3 in front of the 1-3sin..

Answer 19

yep ok so that is important for later on when we apply the chain rule. Now, we need to rewrite a little for this to make sense

Answer 20

GL ... tag me if ya need anything

Answer 21

\[y(\theta)= \theta+(cos(u))^2\]

Answer 22

then I like to do a second substitution here

Answer 23

let's say h=cosu what is dh=?

Answer 24

do you always put the ^2 outside of a trig function when it says cos^2. sin^2, etc…?

Answer 25

dh= -sin(u)

Answer 26

There are 3 chains.
\(\LARGE\color{black}{y=\theta+\cos^2(3\theta )}\)
re-writing the function as,
\(\LARGE\color{black}{y=\theta+[~\cos(3\theta )~]^2}\)
(finding dy / d theta)
\(\LARGE\color{black}{y=1+2[~\cos(3\theta )~]^{2-1}\times[-\sin(3\theta )]\times \theta}\)
\(\LARGE\color{black}{y=1-2\theta\cos(3\theta )\sin(3\theta )}\)

Answer 27

yes, it is just another way to write it.

Answer 28

And you know that:
\(\LARGE\color{black}{\sin(2a)=2\sin(a)\cos(a)}\)

Answer 29

So now, we have \[y(\theta)= \theta+h^2\]

Answer 30

oh writing it that way make much more sense lol now i see how theres multiple chain rules going on

Answer 31

can you find the derivative of that?

Answer 32

so now its 1+2h

Answer 33

yea, that's why I think the substitution makes life easier. It's an extra step but it works

Answer 34

ok so key thing here is that it y'=1+2h*dh*du

Answer 35

that is your chain rule in work

Answer 36

No, we don't need substitution. Do we?

Answer 37

because the total derivative of h requires the previous derivatives too,(why we found them)

Answer 38

ohh, I made an error.

Answer 39

the derivative of 3theta is 3, and I wrote theta-:(

Answer 40

sorry.

Answer 41

so now, can you substitute everything in and tell me what you get @anarvizo

Answer 42

@SolomonZelman , I like doing the substitutions because it makes you a lot less likely to mess up. I think They are a nice in between step that prevents little mistakes.

Answer 43

I made a mistake not because of the method, but because of me:)

Answer 44

okay so we have y'= 1+2(cos(theta))*-sin(theta)*3

Answer 45

I wasn't implying you, sorry, I mean for everyone me included. I feel that way

Answer 46

what was u again?

Answer 47

theta? right?

Answer 48

no u=3theta

Answer 49

yes, that is correct, but one mistake.
y'= 1+2(cos(\(\large\color{black}{ 3 }\)theta))*-sin(\(\large\color{black}{ 3 }\)theta)*3

Answer 50

if you scroll up, you'll see it

Answer 51

ohh duhhhh lol silly me

Answer 52

mistakes happen to all people. you have just now wintessed this.

Answer 53

lol it's ok, but so now that is right, I'll let Solomon explain the double angle simplification. Since he already did it. also wiNTessed? :)

Answer 54

yeah, I am like dude... when I wrote it.

Answer 55

how does that equal to 1-3sin(6theta) though? I have an extra cos unless theres an error or the answer document? hahaha and its fine :) human beings ar ethe best at making mistakes

Answer 56

on*

Answer 57

no no you are correct that answer is correct

Answer 58

it's just one more step to make it pretty

Answer 59

we are applying. \(\large\color{black}{ \sin(2\theta)=2\sin(\theta)\cos(\theta)}\)

Answer 60

okay :)

Answer 61

which in your case is, \(\large\color{black}{ \sin(6\theta)=2\sin(3\theta)\cos(3\theta)}\)

Answer 62

ohh i see where you're going

Answer 63

you can always prove this identity,
it is just sin(a)cos(b)+sin(b)cos(a) where a and b are same.

Answer 64

and after, you know that:
sin(X)cos(X)+sin(X)cos(X)=2sin(X)cos(X).

Answer 65

alrighty

Answer 66

So it is just applying a known formula to make the answer pretty, both answers are equivalent though

Answer 67

yeah, but the one with and angle of 6theta is shorter to write.

Answer 68

true :) but I never memorized those formulas haha never needed to

Answer 69

i had them memorized in precalc; not much anymore :/

Answer 70

I never took pre-calc.... I skipped it

Answer 71

I memorize everything that seems logical. I think double angles for sin and cos are good.

Answer 72

smart..

Answer 73

I just didn't use them enough to warrant memorizing them.

Answer 74

i did it in high school :S I'm a freshman in college lol and yes very smart

Answer 75

I did calc 1 and 2 in hs, we actually didn't have a pre-algebra class. just trig

Answer 76

oh wow! AP calc was the only higher math course offered at my hs 0.o my teacher wasn't all that great though so here i am… lol. thanks for helping though!! all of you :)

Answer 77

If you ever get confused or want to get ahead, what saved me in those was khanacademy.org and MITopencourseware. You can watch the lectures for the subject for free and get a better teacher. Khan is good if you are super confused, MIT is good for if you really understand the last section covered.

Answer 78

np, and be warned college professors can be [extraordinarily]worse than hs, so try to prep for math classes. either watch the lectures or read the book if you understand it, that way during class you're not just like....uhhhmmm what's with all this greek? :)

Answer 79

I've come to notice that about the professors D: lol thanks! haha quick question though; say i have to find where that slope =1… i'd just set the derivative equal to 1 and solve for theta..? I feel like its wrong though cause shouldn't there be multiple answers..?

Answer 80

@FibonacciChick666

Answer 81

I would think so:)

Answer 82

knowing that the derivative is the slope.

Answer 83

okay yeah that what made sense to me; thanks! wanted to make sure:)

Answer 84

this is (similar to) what we do by a mean value theorem.

Answer 85

I agree

Answer 86

and MVT isn't covered until theoretical calc Solomon. I doubt they know it yet

Answer 87

I covered it in Calc 1.

Answer 88

Like I hate how they make separate course/semester for "abstract calc". it is now much different conceptually, besides that any problem would be very not-used to.

Answer 89

i mean not much.

Answer 90

the mean value theorem? we covered it :)

Answer 91

lol, I also thought so:)

Answer 92

really... hmm I don't remember doing it lol

Answer 93

but then again, long time ago haha

Answer 94

I haven't even heard of "theoretical calc"....

Answer 95

.... I just passed it. It is awful. You have to prove everything. EVERYTHING from 1 variable calculus. In one semester

Answer 96

So you prove limits, derivatives, integrals, but there is soooooo much info in between

Answer 97

have you heard of \(\epsilon-\delta\) definition of a limit?