How do you differentiate #e^y cos(x) = 6 + sin(xy)#?

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Answer 1 · 2018-04-17T15:39:08

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

Here,

#e^ycosx=6+sin(xy)#

Diff.w.r.t. #x#,#"using "color(blue)"product and chain rule"#

#e^yd/(dx)(cosx)+cosxd/(dx)(e^y)=0+cos(xy)d/(dx)(xy)#

#e^y(-sinx)+cosx(e^y(dy)/(dx))=cos(xy)[x(dy)/(dx)+y*1]#

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

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

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

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