Question

What is the derivative of `xcos(y) +ycos(x) = 1`?

Answer 1

`dy/dx=(ysinx-cosy)/(cosx-xsiny)`

Explanation:

`xcosy+ycosx=1=> (d(xcosy))/dx+(d(ycosx))/dx=(d(1))/dx=>`

`dx/dxcosy+x(d(cosy))/dx+dy/dxcosx+y(d(cosx))/dx=0=>`

`cosy-xsin(y)dy/dx+dy/dxcosx-ysinx=0=>`

`dy/dx(cosx-xsiny)=ysinx-cosy=>`

`dy/dx=(ysinx-cosy)/(cosx-xsiny)`