Question

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

Answer 1

Given: 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 xx 1 - x xx dy/dx)/y^2`

⇒ `y^2 = xy - x^2 xx dy/dx`

⇒ `x^2 xx dy/dx = xy - y^2`

⇒ `dy/dx = (y(x - y))/x^2`

⇒ `dy/d = y/x xx ((x - y)/x)`

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

`dy/dx = (x - y)/(xlogx)`

Hence proved.