Respuesta :

Edufirst
Assuming that the given expression is    [tex]ln(x+y)= e^{x/y} [/tex]

You calculate dy / dx = y' as:

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

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

Answer dy / dx = - e^(x/y) (x/y^2) (x+y)