Difficult derivitave Please help me!!!?

Question

Calculate dy/dx of Ln(x+y)=e^y/x Please show me how to figure this out! i been on this question for days!! thanks everyone ;-)

Answer 1

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

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

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

Answer 2

Easiest way is technology (Maple output):

> implicitdiff(log(x+y)=exp(y)/x,y,x);
2
x + exp(y) y + exp(y) x
----------------------------
x (-x + exp(y) x + exp(y) y)

Apparently yahoo answers doesnt operate under WYSIWYG, anyway.....

When you differentiate, the resulting equation should be linear in y' and so easy to solve. Lets write it is ln(x+y)=e^y*x^-1 so we can use the product rule. Then d/dx this to get 1/(x+y)*(x+y)'= e^y*y'*x^-1+ e^y*-1*x^-2 so 1/(x+y)*(1+y') = e^y*y'*x^-1+ e^y*-1*x^-2 then you just solve for y'. Imagine y and x and e^y etc. were just numbers like 1 and 2 etc. You would have the equation 1/(1+2)*(1+y')= 3*y'*4 +5*-1*6 and then you would solve that for y'. Similarly solve the equation with the x's and stuff for y' to get the answer above. Good luck.