Question

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)

Answer 1

Hang on ... I'm working on it.

Answer 2

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.

Answer 3

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

Answer 4

hmm, i just realized my mistake. i forgot about the chain rule on cot(xy)

Answer 5

\[y'= - \cot(xy)/x^2\]

Answer 6

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.

Answer 7

\[xy=\cot(xy)\]
\[xy' + y=-\csc^2(xy) (xy)'\]
\[xy' + y=-\csc^2(xy) (xy' + y)\]
\[xy'(1+\cos^2(xy))=-y(1+\cos^2(xy))\]
\[xy'=-y\]
\[y'=-y/x\]
we know that \[y = \cot(xy) / x\]
replace it
\[y' =- \cot(xy) /x^2\]

Answer 8

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?

Answer 9

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.

Answer 10

sorry 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!

Answer 11

Please let me know if you understand?

Answer 12

@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 \[y' = [-y \csc^2(xy)-y]/[x + x \csc^2(xy)]\]