How do you find #dy/dx# if #cosy=e^(x+y)#?

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Answer 1 · 2018-02-23T08:24:33

#dy/dx=-(e^xe^y)/(e^xe^y+sin(y))#

We want to #dy/dx# when

#cos(y)=e^(x+y)#

Differentiate the left side by the chain rule

#LHS=d/dxcos(y)=-dy/dxsin(y)#

Differentiate the right side by the chain and product rule

#RHS=d/dxe^(x+y)=d/dxe^xe^y=e^xe^y+dy/dxe^xe^y#

These sides should equal

#-dy/dxsin(y)=e^xe^y+dy/dxe^xe^y#

#=>dy/dxe^xe^y+dy/dxsin(y)=-e^xe^y#

#=>dy/dx(e^xe^y+sin(y))=-e^xe^y#

#=>dy/dx=-(e^xe^y)/(e^xe^y+sin(y))#