Question

If x^y = e^{x – y} prove that `dy/dx = logx/((log(xe))^2` and hence find its value at x = e.

Answer 1

Given, x^y = e^{x – y}

Taking log of both sides:

log (x^y) = log(e^{x – y})

y × log x = (x – y) log e  ....(As log (a^b) = b ⋅ log a)

y × log x = (x − y)  ...(Since log e = 1)

Differentiating both sides w.r.t. x.

`(d(y xx log x))/dx = (d(x - y))/dx`

Using the product rule:

As (uv)' = u'v + v'u

`(d(y))/dx * log x + (d(log x))/dx * y = (d(x))/dx - (d(y))/dx`

`(dy)/dx log x + 1/x * y = 1 - (dy)/dx`

`(dy)/dx log x + y/x = 1 - dy/dx`

`(dy)/dx log x + dy/dx = 1 - y/x`

`dy/dx (log x + 1) = ((x - y)/x)`

Since y = log = (x − y)

y log x + y = x

x = y log x + y

x = y(1 + log x)

`dy/dx (log x + 1) = (y(1 + log x) - y)/(y(1 + log x))`

`dy/dx (log x + 1) = (y + y log x - y)/(y(1 + log x))`

`dy/dx (log x + 1) = (y log x)/(y(1 + log x))`

`dy/dx (log x + 1) = (log x )/(1 + log x)`

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

Putting log e = 1

`dy/dx = (log x)/(log e + log x)^2`

Using log A + log B = log (AB)

`dy/dx = (log x)/(log (xe))^2`

Hence proved.

Now, we need to find the value of `dy/dx` at x = e.

Putting x = e in `dy/dx`:

`dy/dx = (log e)/(log (e xx e))^2`

`dy/dx = (log e)/(log (e^2))^2`

Using log Aⁿ = n × log A

`dy/dx = (log e)/(2 xx log e)^2`

Putting log e = 1

`dy/dx = 1/(2 xx 1)^2`

`dy/dx = 1/2^2`

`dy/dx = 1/4`