Calculate dy/dx if Ln (x + y) = ex/y A: e^x/y xy + e^x/y y^2 - y^2 ______________________ e^x/y xy + e^x/y x^2 + y2B: e^x/y xy + e^x/y y^2 - y^2 ______________________ e^x/y xy - e^x/y x^2 + y^2C: e^x/y xy + e^x/y y^2 + y^2 ______________________ e^x/y xy - e^x/y x^2 + y^2D: e^x/y xy + e^x/y y^2 + y^2 ______________________ e^x/y xy + e^x/y x^2 + y^2

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

In this case we use the quotient rule on the right hand side where we have an exponent.

1/(x+y)[1+dy/dx]=e^(x/y)[[(y)-x(dy/dx)]/[y^2]]

opening the parenthesis on the RHS

1/(x+y)[1+dy/dx]=e^(x/y)[1/y-(x/y^2)(dy/dx)]

Expanding the left hand side and right hand side
 
1/(x+y)+(dy/dx)[1/(x+y)]=[e^(x/y)]/y-[e^(x/y)](x/y^2)(dy/dx)

next we put like terms together by moving all terms with dy/dx to the LHS and everything else to the RHS.

(dy/dx)[1/(x+y)]+[e^(x/y)](x/y^2)(dy/dx)=[e^(x/y)]/y-1/(x+y)


dy/dx[1/(x+y)+[e^(x/y)](x/y^2)]=[e^(x/y)]/y-1/(x+y)

we can replace e^(x/y) with ln(x+y)

dy/dx[1/(x+y)+[ln(x+y)](x/y^2)]=[ln(x+y)]/y-1/(x+y)

Multiplying both sides by (x+y)(y^2)

dy/dx[y^2+(x+y)ln(x+y)]=y(x+y)ln(x+y)-y^2

dy/dx=[y(x+y)ln(x+y)-y^2]/[y^2+(x+y)ln(x+y)]

This can be written as:
dy/dx=[y(x+y)e^(x/y)-y^2]/[y^2+(x+y)e^(x/y)]
the answer is A]