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

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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!

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