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)
Stacey Warren - Expert brainly.com
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Hang on ... I'm working on it.
Okay, I got a very different answer. Why don't you write out your first couple of lines so I can see what you did.
using '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
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hmm, i just realized my mistake. i forgot about the chain rule on cot(xy)
\[y'= - \cot(xy)/x^2\]
Sorry 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.
\[xy' + y=-\csc^2(xy) (xy)'\]
\[xy' + y=-\csc^2(xy) (xy' + y)\]
we know that \[y = \cot(xy) / x\]
\[y' =- \cot(xy) /x^2\]
thanks 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?
Sorry 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.
sorry for the mistake:
Then the others steps are right!
Please let me know if you understand?
@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 \[y' = [-y \csc^2(xy)-y]/[x + x \csc^2(xy)]\]