## 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)

1. anonymous

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

2. anonymous

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.

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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.

7. anonymous

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

8. anonymous

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?

9. anonymous

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.

10. anonymous

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!

11. anonymous

Please let me know if you understand?

12. anonymous

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