How do you differentiate (e^y)(cos(x))=1+sin(xy)?

1 Answer
May 11, 2018

dy/dx={ycos(xy)+e^ysinx}/{e^ycosx-xcos(xy)}.

Explanation:

Given that, e^ycosx=1+sin(xy).

:. d/dx{e^ycosx}=d/dx{1+sin(xy)}.

Applying the Product Rule, and, the Chain Rule, we have,

:. e^y*d/dx(cosx)+cosx*d/dx(e^y)={0+cos(xy)}*d/dx(xy).

:. -e^ysinx+(cosx){d/dy(e^y)}dy/dx=(cos(xy)){x*d/dx(y)+y*d/dx(x)}.

:. -e^ysinx+e^ycosxdy/dx=xcos(xy)dy/dx+ycos(xy).

:. e^ycosxdy/dx-xcos(xy)dy/dx=ycos(xy)+e^ysinx.

:. {e^ycosx-xcos(xy)}dy/dx=ycos(xy)+e^ysinx.

:. dy/dx={ycos(xy)+e^ysinx}/{e^ycosx-xcos(xy)},

as desired!