Question

If x = `"e"^(x/y)`, prove that `"dy"/"dx" = (x - y)/(xlogx)`

Answer 1

Given that: x = `"e"^(x/y)`

Taking log on both the sides,

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

⇒ log x = `x/y log "e"`

⇒ log x = `x/y`  ....[∵ log e = 1]  ....(i)

Differentiating both sides w.r.t. x

`"d"/"dx" log x = "d"/"dx"(x/y)`

⇒ `1/x = (y*1 - x* "dy"/"dx")/y^2`

⇒ y² = `xy - x^2 * "dy"/"dx"`

⇒ `x^2 * "dy"/"dx"` = xy – y²

⇒ `"dy"/"dx" = (y(x - y))/x^2`

⇒ `"dy"/"d" = y/x * ((x - y)/x)`

⇒ `"dy"/"dx" = 1/logx * ((x - y)/x)`  .....[∵ log x = `x/y` from equation (i)]

Hence, `"dy"/"dx" = (x - y)/(xlogx).