Diferentiate y=arctan(cos(theta))

Nah, you're good, dude. You just have to use a little bit of intuition. The fact that you realize you could differentiate in terms of y or in terms of theta and get different answers is a good sign. But you just have to realize the question wants you to differentiate in terms of theta. A better wording of the question would have been "y=f(theta) is a function of theta. Differentiate y."

Thanks for the help guys. Does this make sense or am I still just confused about what the question even is asking?

sorry idk why it turned sideways it was correct on my computer... maybe this crop fixed it?

it's correct except for the fact that you should put a parenthesis around -sin to show it's multiplying

You shouldn't be worried. Everyone in this thread is retarded. You are correct the question never tells you whether to find dy/dtheta or dtheta/dy. But everyone else saying the variable doesn't matter is just confusing you. What they probably meant was that looking for a derivative is wrong whether it's dy/dtheta or dtheta/dy. The problem told you to differentiate, this is different from finding a derivative. Although it's a subtle language difference, it can cause lots of confusion. When finding a derivative you need to do it with respect to something. Differentiation is just doing the act of taking a derivative using every variable. For example, your problem was y=arctan(cos(theta)). Now you can differentiate both sides: dy=(1/(1+cos^2(theta)))*-sin(theta)*dtheta. Notice that from here you can find dy/dtheta but it's not necessary because the question just asked you to differentiate.

>Now you can differentiate both sides: dy=(1/(1+cos^2(theta)))*-sin(theta)*dtheta
>you can multiply and divide differentials
jesus fucking christ how heretic
please stop

>has never solved for a derivative implicitly

Wow, thank you that makes so much more sense.

You have to differentiate both sides... with respect to a new variable (eg. t), right?

I honestly think it's just a badly worded question, or OP didn't give the fully worded version. It can't be that complicated for fucking high school shit.

if you want to be technical about notation, then "dy/dθ y" is repetitive; you could say "dy/dθ" or just "d/dθ y" or "d f(θ)/d(θ)" assuming that y=f(θ)

also what said

stop. this method is wrong and doesn't make sense. you can't manipulate an operator like that and split it in two. there's a reason mathematicians have to learn this shit all over again doing it properly and call it analysis

Yes, good call. I'm not sure how i dropped the parentheses.

>there's a reason mathematicians have to learn this shit all over again doing it properly and call it analysis
And what reason is that?

that calculus is unrigorous shit that pulls of this kind of objectively incorrect tricks, and people eat it right up because it's one more unexplained magic trick that they promise works
splitting a fucking operator in half, holy fuck

You keep saying I'm splitting an operator in half...are you referring to dy/dtheta? I don't know what your native language is but that's a derivative not an operator. And if you are talking about some other operator can you clarify what it is before you continue spouting shit and ranting about your useless pure math degree?

i'm referring to the d/dtheta derivative operator that you're applying to both sides, you stupid fucking baboon

>that's a derivative not an operator
wow

Alright fair enough I'm retarded. But that still doesn't answer the question. Why exactly is splitting the operator in half an objectively incorrect trick in your mind? I still came out with the right answer. I understand that you think I'm an unrigorous pleb but why isn't what I did valid? What's wrong with saying: " the change in y is equal to -sin(theta)/(1+cos^2(theta)) times the change in theta." Why does change have to be with respect to something else? If this problem had 3 variables we could make a similar statement without having to worry about partial derivative notation. Seems perfectly rigorous to me

>If this problem had 3 variables we could make a similar statement without having to worry about partial derivative notation
no. let me show you how you can quickly fuck shit up
define x = yz (what I mean, let f be the function from R^2 to R defined by multiplication. implicitly we call this function x)
dx = z dy + y dz (what do I mean? nothing, I'm applying a retarded notational trick. what are both sides? what does dx mean? nothing. it doesn't make sense anymore)
dx/dy = z + y dz/dy (again, another retarded trick, divide by dy. what is this shit even? this is not the partial derivative of our function with respect to the first projection like dx/dy implies, because that's z, the second projection)

...

jfc autists

>dx = z dy + y dz (what do I mean? nothing, I'm applying a retarded notational trick. what are both sides? what does dx mean? nothing. it doesn't make sense anymore)
>what do I mean?
You mean the change in x is z times the change in y plus y times the change in z.
>what does dx mean?
The change in x
>dx/dy = z + y dz/dy (again, another retarded trick, divide by dy. what is this shit even? this is not the partial derivative of our function with respect to the first projection like dx/dy implies, because that's z, the second projection)
>this is not the partial derivative of our function with respect to the first projection like dx/dy implies, because that's z, the second projection)
Except it is the partial derivative dx/dy. You just need to assume the change in z is 0 (dz=0). This is what you're supposed to assume anyway when talking about partials. The form you have written it in is NOT unrigorous, it just requires a few extra unspoken rules (Every notation has this, it's unavoidable). In fact, I'd argue that the form you've written it in shows more underlying information. For example, what if dz does not equal 0? You claim that dx/dy is nonsense unless dz is 0, but I think that's close-minded. What if it isn't nonsense? What if our brains just can't comprehend dx/dy when dz=1? Don't call me a retard, all I'm saying is my way of writing it is equivalent to yours, there is just some extra stuff in there that might confuse the uneducated.

>You mean the change in x is z times the change in y plus y times the change in z.
this doesn't mean anything. those are buzzwords with no substance. write this as actual mathematics if you can, and you'll realize you can't. the first expression made much more sense because I knew what it was: the function from R^2 to R defined by multiplication.

>Except it is the partial derivative dx/dy when you fuck up the numbers for it to kind of work
dz = 0 implies dx/dz = dx/0 and that's undefined. that's nonsense as well, dz is supposed to be "the change in z", whatever that means. it's suddenly 0 while "the change in y" is not? the function is symmetric.

really, at this point you don't want to understand, all you want to do is justify ridiculous abuses of notation. I'm not going to reply unless you write mathematically what dx means. as an example, what d/dx usually means is the operator that takes a function f:R^n -> R^m to the function lim(h->0)( f(x1+h,x2,x3,...,xn) - f(x1, x2, x3, ..., xn) )/ h