Question

Find `"dy"/"dx"` , if `"y" = "x"^("e"^"x")`

Answer 1

`"y" = "x"^("e"^"x")`

Taking logarithm of both the sides, 

log y = e^x log x

Differentiating both sides w.r.t. x, 

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

`= "e"^"x". 1/"x" + "log x". "e"^"x"`

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

`therefore "dy"/"dx" = "y".["e"^"x" (1/"x" + "log x")]`

`= "x"^("e"^"x") . "e"^"x" (1/"x" + "log x")`