Nasty Implicit differentiation: find dx/dy for xy = cot(xy)

[scrum]

These are my homework problems that are dude tomrrow that I'm just giving up on. I just thought I'd post to see what they are.

1) Find dx/dy by implicit differentiation.

xy = cot(xy)

I get

(xy'+y)=csc(xy)^2 * (xy'+y)

and then the (xy'+y) cancel and I get

0 = csc(xy)^2 but that is useless to me. I don't really know where i'm going wrong.

2) If g(x) + x sin g(x) = x^2, and g(1) = 0, find g'(1).

this I had no idea how to approach. I know it doesn't work at g(1) but I don't know where to go form there.
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Edited by stapel -- Reason for edit: Replacing graphic with text.

 

[o_O]

1. Well you don't get 0 = -csc²(xy), you get 1 = -csc²(xy). Either than that, instead of cancelling just multiply through and solve for y'.

Edit: Oh ya. (cotx)' = -csc²x. You forgot the negative sign.

2. Would it be easier if you imagined y = g(x)? y + xsiny = x². Just implicit differentiate and plug, plug, plug.

 

[tkhunny]

Re: Nasty Implicit differentiation: find dy/dx for xy = cot(

1) Find dx/dy by implicit differentiation.
xy = cot(xy)

Did you notice what a strange equation this is? Does such a derivative actually exist? A quick examination of x = cot(x) doesn't show much in a single plane.

Just a thought question.

 

[Deleted member 4993]

y' is dy/dx - however you have been asked to find dx/dy - isn't it.

Very interesting problem - show some more work using the above hints.

 

[daon]

\(\displaystyle y = \frac{cot(xy)}{x}\)

\(\displaystyle dy = \frac{-xcsc^2(xy)(ydx+xdy) - cot(xy)dx}{x^2}\)

\(\displaystyle \frac{dy}{dx} = \frac{-xcsc^2(xy)(y+x\frac{dy}{dx}) - cot(xy)}{x^2}\)

\(\displaystyle \frac{dy}{dx} = \frac{-xcsc^2(xy)y-x^2csc^2(xy)\frac{dy}{dx} - cot(xy)}{x^2}\)

\(\displaystyle \frac{dy}{dx} = \frac{-xcsc^2(xy)y}{x^2}- csc^2(xy)\frac{dy}{dx} - \frac{cot(xy)}{x^2}\)

\(\displaystyle \frac{dy}{dx}\(1+ csc^2(xy)\) = \frac{-csc^2(xy)y}{x} - \frac{cot(xy)}{x^2}\)

\(\displaystyle \frac{dy}{dx} = \frac{-xcsc^2(xy)y - cot(xy)}{x^2\(1+ csc^2(xy)\)}\)

Or comming from what you started to do, one would solve and get:

\(\displaystyle y'=\frac{-csc(xy)^2y-y}{x\(1+csc^2(xy)\)}\)

They are equivilant answers. To see this multiply by \(\displaystyle \frac{x}{x}\). The second term in the numerator is xy, but by assumption xy=cot(xy).