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anonymous
 5 years ago
just checking if i got this right, differentiate implicitly:
xy = cot(xy)
i went through the steps and got:
y' = (cot y  csc^2(xy)  y)/(x  cot x)
anonymous
 5 years ago
just checking if i got this right, differentiate implicitly: xy = cot(xy) i went through the steps and got: y' = (cot y  csc^2(xy)  y)/(x  cot x)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hang on ... I'm working on it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I got a very different answer. Why don't you write out your first couple of lines so I can see what you did.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using 'D' to represent derivative: x D(y) + y D(x) = cot D(xy) + (xy) D(cot) xy' + y = cot(xy' + y)  csc^2 (xy) distributed the cot over (xy' + y) and isolated y' to get my answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, i just realized my mistake. i forgot about the chain rule on cot(xy)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y'=  \cot(xy)/x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry for the delay  system froze/crashed ... again! Yes  remember, D(cot x) is csc^2(x). And there should be no cot in the answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[xy=\cot(xy)\] \[xy' + y=\csc^2(xy) (xy)'\] \[xy' + y=\csc^2(xy) (xy' + y)\] \[xy'(1+\cos^2(xy))=y(1+\cos^2(xy))\] \[xy'=y\] \[y'=y/x\] we know that \[y = \cot(xy) / x\] replace it \[y' = \cot(xy) /x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks guys for the help! i understand up to \[xy' + y = \csc^2(xy)(xy' + y)\] but i'm not sure how to manipulate it to isolate y' on it's own. @corec was that a trig identity you're using?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry for the long delay  this web site crashes/hangs/freezes on me regularly and I was unable to get back in last night. You're right  after the line you understood, an error was made. Somehow, csc was mixed up with cos. The change was an error, not a trig identity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for the mistake: \[xy′(1+\cos^2(xy))=−y(1+\cos^2(xy)) \] replace with \[xy′(1+\csc^2(xy))=−y(1+\csc^2(xy)) \] Then the others steps are right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please let me know if you understand?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@corec finally got it. i didn't think to distribute the \[\csc^2(xy)\] over \[(xy' + y)\] to isolate the y'. so i came out with \[y' = [y \csc^2(xy)y]/[x + x \csc^2(xy)]\]
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