What is 2xy differentiated implicitly?

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Answer 1 · 2015-09-02T18:31:31

#y'=(2y)/(1-2x)#

The question does not specify with respect to what so I'll assume y is a function of x.

#y' = d((u.v))/dx=v.du/dx+u.dv/dx#

So:

#y'=2x.y'+y2.dx/dx#

#y'=2x.y'+2y#

#y'=(2y)/(1-2x)#

Answer 2 · 2018-07-20T14:43:59

The answer is #=-y/x#

The function is

#f(x,y)=2xy#

The partial derivatives are

#(delf)/(delx)=2y#

#(delf)/(dely)=2x#

Therefore,

#dy/dx=-((delf)/(delx))/((delf)/(dely))=-(2y)/(2x)=-y/x#