Question

x^y = e^x-y, then show that  `"dy"/"dx" = ("log  x")/("1 + log x")^2`

Answer 1

x^y = e^x-y 

Taking logarithm of both the sides 

y log x = (x - y) log e

            = x - y           ...(1)

Diff. both the sides w.r.t. x , 

`"y".1/"x" + "log x" . "dy"/"dx" = 1 - "dy"/"dx"`

`therefore "log x" . "dy"/"dx" + "dy"/"dx" = 1 - "y"/"x"`

`therefore (1 + "log x") "dy"/"dx" = ("x - y")/"x"`

`therefore (1 + "log x") "dy"/dx"= (ylogx)/x`   ...[by (1)]

`"dy"/"dx" = ("y log x")/("x" . (1 + "log x"))`

`= ("log x")/(1 + "log x")^2`

                      `[because "from (1)"  "y"/"x" = 1/(1 + "log x")]`