Math help: Find dy/dx if [sin(xy)]/(2x-3y)=y?

Question

I can't remember what to do with the y in the equation. Do I treat it like a constant, or do I have to differentiate with respect to x and then with respect to y? That sonds familiar, but it's been so long, I can't remember.
Thanks

Answer 1

Implicit differentiation:

((2x - 3y)cos(xy)(xdy + ydx) - sin(xy)(2dx - 3dy)/(2x - 3y)^2 = dy
(2x - 3y)cos(xy)(xdy + ydx) - sin(xy)(2dx - 3dy) = dy*(2x - 3y)^2

dy((2x - 3y)xcos(xy) + 3sin(xy)) + dx((2x - 3y)ycos(xy) - 2sin(xy)) = dy*(2x - 3y)^2

solve for dy/dx

Answer 2

Dear
To be clear Y consider to be constant only if we speak of partial integration which could not be the case as the equation has two variable only.

Let us start
1- Integrate both Sides with respect to x .
2- R.H.S will be = dy/dx call it Dy
3- L.H.S will be
[cos(xy)(2x-6y)(y+Dy)- sin(xy)(2-3Dy)]/(2x-3y)^2

that it is just solve it for Dy
If you Have any comment just send it to me

Answer 3

Usually in problems like this, the x is considered an independent variable, and the y a function of x. To find dy/dx = y', just differentiate both sides with respect to x, keeping in mind that y is a function of x, so the chain rule applies:

cos(xy)(y + xy')(2x - 3y) - sin(xy)(2 - 3y')
-------------------------------------------------------------------------- = y'
(2x - 3y)^2

Now solve this for y', and that's the answer.

Answer 4

This is a nasty equation!

For the left side, use quotient rule:

(cos(xy)(x dy/dx+ y) is the derivative of sin(xy) by chain and product rule

( (2x-3y)(cos(xy)(x dy/dx+ y) - sin(xy) (2-3 dy/dx) ) / (2x-3y)^2 = dy/dx

Now solve for dy/dx!