Answer:
See below.
Explanation:
1. x = e^(x/y) Taking logs:
log x = x/y
x = y log x
Differentiating:
1 = dy/dx log x + y * 1/x
dy/dx log x = (1 - y/x)
dy/dx = (1 - y/x) / log x
dy/dx = ((x - y)/ x) / log x
dy/dx = (x - y) / x log x)
2. y^x = e ^(y - x)
Taking logs:
x log y = y - x --------------(A)
y = x + x logy --------------(B)
dy/dx = 1 + x * 1/y * dy/dx + 1*logy
dy/dx - dy/dx * x/y = 1 + log y
dy/dx ( (x - y) y)) = 1 + log y
dy/dx = y(1 + log y) / (y - x)
Using (A) and (B) :-
dy/dx = x(1 + logy)(1 + logy) / x logy The x's cancel, so:
dy/dx = (1 + logy)^2 / log y
Sorry I have to go right now..
The rest of the answers are on the re-post of these questions