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))#