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

1 Answer

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

Explanation:

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.