Implicit differentiation?

Question

y=sin(xy) with respect to y.

i got here: y'=cos(xy)(y+y'x)

and now im stuck

Answer 1

so far, so good ... [ if you want derivatives with respect to x, that is ] ... If you really want the derivatives done wrt to y, see my EDIT below ......

now multiply out the right side to get ....

y ' = ycos(xy) + y ' ( xcos(xy))

and collect like terms...

y ' - y ' ( xcos(xy)) = ycos(xy), factor ,

y ' { 1 - xcos(xy) } = y cos( xy )

and now divide ...

so y ' = y cos( xy ) / { 1 - xcos(xy) }, where ... y ' = dy/dx

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EDIT ... AbeLincoln below is correct on taking derivatives with respect to y, as you seem to be asking in your question, but his answer is incorrect .......

if you want the derivative wrt to y, then it is
1 = cos( xy ) * [ x + yx ' ], where x ' = dx/dy

then x ' = [ 1 - xcos(xy) ] / ( ycos( xy ) )

or x ' = ( 1 / (ycos ( xy ) ) - ( x / y )


good job in catching the derivative wrt y part , Abe..... but there is an error in your work after that .... see my work with EDIT...

Answer 2

No it is not y' = cos(xy)(y + y'x)

You are taking the derivative with respect to y so d/dy (y) = 1

The right side of the equation you have is correct, when you do regular algebraic manipulations you will get:

dx/dy = [ 1/(cos(xy) + x](1/y)

Answer 3

Ok that's right.

y' =cos(xy)*y +cos(xy)*x*y'

y' *(1 -cos(xy)*x) = y*cos(xy)

y' = y*cos(xy)/(1-x*cos(xy))

Answer 4

now that you have y'...subsitute