Calculate dy/dx if Ln(x+y)=e^x/y

Answer 1

This is an add on to the question above. Thought this might help here are the answer choices:
A. e^x/y xy+e^x/y y^2+y^2
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e^x/y xy+e^x/y x^2+y^2

B. e^x/y xy+e^x/y y^2-y^2
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e^x/y xy-e^x/y x^2+y^2

C. e^x/y xy+e^x/y y^2+y^2
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e^x/y xy-e^x/y x^2+y^2

D. e^x/y xy+e^x/y y^2-y^2
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e^x/y xy+e^x/y x^2+y^2

Hope these will help!

Answer 2

just work it with implicit differentiation:

ln(x+y)=e^x/y

1/(x+y) * (1 + y') = e^x/y (1/y - x/y^2 y')

1/(x+y) + 1/(x+y) y' = 1/y e^x/y - e^x/y xy'/y^2

(1/(x+y) + e^x/y x/y^2)y' = 1/y e^x/y - 1/(x+y)

Putting it all over (x+y)y^2 we have

[y(x+y)e^x/y - y^2]/[y^2 + x(x+y)e^x/y]
= D

Answer 3

calculate dy/dx when y=xy3+y=3x.