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

1 Answer
Jul 19, 2017

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)