Question

If `x^y = e^(x - y)` , show that `(dy)/(dx) = logx/(1 + logx)^2`

Answer 1

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

Taking log on both sides , 

`log(x)^y = loge^(x - y)`

∴ y . logx = (x - y).loge

∴ y logx = x - y  ...........(∵ loge = 1)

∴ y logx + y = x

∴ y(logx + 1) = x

∴ y = `x/(1 + logx)`

Differentiating w.r.t. x, 

`(dy)/(dx) = ((1 + logx). 1 - x(0 + 1/x))/(1 + logx)^2`

∴ `(dy)/(dx) = (1 + logx - 1)/(1 + logx)^2`

∴ `(dy)/(dx) = logx/(1 + logx)^2`

Hence proved .

Answer 2

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

Taking log on both sides , 

`log(x)^y = loge^(x - y)`

∴ y . logx = (x - y).loge

∴ y logx = x - y  ...........(∵ loge = 1)

∴ y logx + y = x

∴ y(logx + 1) = x

∴ y = `x/(1 + logx)`

Differentiating w.r.t. x, 

`(dy)/(dx) = ((1 + logx). 1 - x(0 + 1/x))/(1 + logx)^2`

∴ `(dy)/(dx) = (1 + logx - 1)/(1 + logx)^2`

∴ `(dy)/(dx) = logx/(1 + logx)^2`

Hence proved .

Answer 3

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

Taking log on both sides , 

`log(x)^y = loge^(x - y)`

∴ y . logx = (x - y).loge

∴ y logx = x - y  ...........(∵ loge = 1)

∴ y logx + y = x

∴ y(logx + 1) = x

∴ y = `x/(1 + logx)`

Differentiating w.r.t. x, 

`(dy)/(dx) = ((1 + logx). 1 - x(0 + 1/x))/(1 + logx)^2`

∴ `(dy)/(dx) = (1 + logx - 1)/(1 + logx)^2`

∴ `(dy)/(dx) = logx/(1 + logx)^2`

Hence proved .