Question

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

Answer 1

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

Taking logarithm of both sides, we get

y log x = (x - y) log e = x - y

∴ y log x + y = x

∴ y(1 + log x) = x

∴ y = `"x"/(1 + log "x")`

Differentiating both sides w.r.t. x, we get

`"dy"/"dx" = "d"/"dx" ["x"/(1 + log "x")]`

∴ `"dy"/"dx" = ((1 + log "x") "d"/"dx" ("x") - "x" "d"/"dx" (1 + log "x"))/(1 + log "x")^2`

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

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

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