What is the derivative of e^y cos x = 3 + sin(xy)?

1 Answer
Mar 28, 2017

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

Explanation:

"differentiate " e^ycosx" using "color(blue)"product rule"

"Given " f(x)=g(x)h(x)" then "

f'(x)=g(x)h'(x)+h(x)g'(x)

"differentiate " sin(xy)" using "color(blue)" chain rule"

"Given " f(x)=g(h(x))" then"

f'(x)=g'(h(x)).h'(x)

"differentiate "color(blue)"implicitly with respect to x"

rArre^y(-sinx)+cosxe^y.dy/dx=cos(xy).d/dx(xy)

rArre^ycosx.dy/dx-e^ysinx=cos(xy)[x.dy/dx+y]

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

rArr dy/dx(e^ycosx-xcos(xy))=e^ysinx+ycos(xy)

rArr dy/dx=(e^ysinx+ycos(xy))/(e^ycosx-xcos(xy))