How do you differentiate #1=e^(xy)/(e^x+xy)#?

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Answer 1 · 2016-11-30T19:13:45

#y'=(e^x+y-ye^x-xy^2)/(xe^x+x^2y-x)#

From the given

#1=e^(xy)/(e^x+xy)#

#e^x+xy=e^(xy)#

#d/dx(e^x)+d/dx(xy)=d/dx(e^(xy))#

#e^x*d/dx(x)+x*d/dx(y)+y*d/dx(x)=e^(xy)*[x*d/dx(y)+y*d/dx(x)]#

#e^x+xy'+y*1=e^(xy)*(xy'+y*1)#

#e^x+xy'+y=e^(xy)*(xy'+y)#

#e^x+y-y*e^(xy)=xy'*e^(xy)-xy'#

#e^x+y-y*e^(xy)=(x*e^(xy)-x)y'#

#y'=(e^x+y-y*e^(xy))/(x*e^(xy)-x)#

But, #e^(xy)=e^x+xy#

therefore

#y'=(e^x+y-y*e^(xy))/(x*e^(xy)-x)=(e^x+y-y*(e^x+xy))/(x*(e^x+xy)-x)#

#y'=(e^x+y-ye^x-xy^2)/(xe^x+x^2y-x)#

God bless....I hope the explanation is useful.