Question

Help me answer these questions

Answer 1

Answer:

See below.

Step-by-step 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