appleme Posted March 16, 2008 Report Share Posted March 16, 2008 How would you finddydxfor 2xy2 = x2y+3 I know that one of the first steps of the working is2y2 + 4xydy = 2xy + x2dy...............dx..................dxI know how to go from there but I dont know how you would get there.. and why? Link to post Share on other sites More sharing options...
Abu Posted March 16, 2008 Report Share Posted March 16, 2008 You get there by using the product rule. I can't answer why , I'm not here to derive the way product rule works Link to post Share on other sites More sharing options...
uggy Posted April 4, 2008 Report Share Posted April 4, 2008 (edited) dy/ dx for the above question is:2xy - 2y^2--------------4xy - x^2The produce rule is: (1st x Differential of the 2nd )+ (2nd x differential of the 1st)Edit: I just saw your working.. Here is how to do the rest.Bring all those with dy/dx to the left hand side..so... 2y^2 + 4xy(dy/dx) = 2xy + x^2(dy/dc)goes to:4xy (dy/dx) - x^2(dy/dx) = 2xy - 2y^2.Then factor out the dy/dx:DY/DX{4xy - x^2} = 2xy - 2y^2Then divide by 4xy - x^2DY/DX = 2xy - 2y^2 -------------- 4xy - x^2 Edited April 4, 2008 by uggy Link to post Share on other sites More sharing options...
ezex Posted April 4, 2008 Report Share Posted April 4, 2008 Do something to make your life easier, do the problem as if it was a regular differentiation, but every time you take the derivative of y just write dy/dx next to it: xy = 2 you dont notice, but when you take the derivative of x the answer isnt 1, its 1*dx/dx which is one. So for xy = 2 just do a product rule: (dx/dx)y + x (dy/dx) = 0 see all you do is put in dy/dx whenever you TAKE THE DERIVATE OF Y. As far as how to do the rest uggy took care of most of it. See in reality what you have is: 2xy2 = x2y+3 (2)(dx/dx)y^2 + 2x(2y)(dy/dx) = (2x)(dx/dx) + 2x(dy/dx) + 0, which is the same as 2y^2 + 4xy(dy/dx) = 2xy+ 2x(dy/dx) Link to post Share on other sites More sharing options...
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