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

Question

11960 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-11T18:15:42

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

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!