Question

If x^y - y^x = a^b, find `(dy)/(dx)`.

Answer 1

The given function is x^y - y^x = a^b
Let x^y = u and y^x = v
Then , the function becomes u - v = a^b

`(d"u")/(d"x") - (d"v")/(d"x") = 0`         .....(1)
u = x^y
⇒ log u = log(x^y)
⇒ log u = y log x

Differentiating both sides with respect to x, we obtain

`(1)/("u") (d"u")/(d"x") = log "x" (d"y")/(d"x") + "y". (d)/(d"x") (log "x")`

⇒ `(d"u")/(d"x") = [ log "x" (d"y")/(d"x") + "y" .(1)/("x")]`

⇒ `(d"u")/(d"x") = "x"^"y" (log "x"  (d"y")/(d"x") + ("y")/("x"))`     ...(2)

`"v" = "y"^"x"`

⇒ `log "v" = log ("y"^"x")`

⇒ `log "v" = "x" log "y"`

Differentiating both sides with respect to x, we obtain

`(1)/("v").(d"v")/(d"x") = log "y".(d)/(d"x") ("x") + "x". (d)/(d"x") (log "y")`

⇒ `(d"v")/(d"x") = "v" (log "y". 1 + "x".(1)/("y"). (d"y")/(d"x"))`

⇒ `(d"v")/(d"x") = "y"^"x" (log "y" + ("x")/("y") (d"y")/(d"x"))`      ...(3)

From (1), (2), and (3), we obtain

`"x"^"y" (log "x"(d"y")/(d"x") + ("y")/("x")) - "y"^"x" (log "y" + ("x")/("y") (d"y")/(d"x")) = 0`

⇒ `"x"^"y" log "x" (d"y")/(d"x") - "xy"^("x"-1) (d"y")/(d"x") + "x"^("y"-1) "y"-"y"^"x" log"y" = 0`

⇒ `("x"^"y" log "x" - "xy"^("x"-1)) (d"y")/(d"x") = "y"^"x" log "y" - "x"^("y"-1) "y"`

⇒ `(d"y")/(d"x") = ("y"^"x" log "y" - "x"^("y" -1) "y")/(("x"^"y" log "x" - "xy"^("x"-1))`