Question

How do you differentiate `f(x)=ln(x)^x-lnx/x^x`?

Answer 1

`y' = {(1 - x^-x)/x} + ln(x){(1+ln(x))(x^-x)}`

Explanation:

Start with assuming `f(x) = y`.

Therefore,

`y = ln(x) - ln(x)/x^x`

`:. y = ln(x)[1 - 1/x^x]`

`:. y = ln(x)[1 - x^-x]`

Now, differentiating wrt x,

`y' = d/dx(ln(x){1 - x^-x})`

`:. y' = (1 - x^-x)d/dxln(x) + ln(x)d/dx(1 - x^-x)` ...(1)

(Using the Chain Rule of Differentation)

The first term is simple to handle, and it reduces to `(1 - x^-x)/x` but the second term is tricky, so we'll evaluate it separately.

Let `t = x^-x`

Taking the natural log on both sides of the equation,

`ln(t) = -xln(x)`

Differentiating both sides wrt x,

`1/tdt/dx = -(1 + ln(x))`

`:. dt/dx = -(1 + ln(x)).t`

Replacing value of t,

`d/dx(x^-x) = -(1 + ln(x))(x^-x)`

Extrapolating further and using the value `1 - x^-x`

`d/dx(1 - x^-x) = (1+ln(x))(x^-x)`

Using the above expression in (1),

`y' = {(1 - x^-x)/x} + ln(x){(1+ln(x))(x^-x)}`

This can be further simplified to suit the provided answer. But this, I believe, is in a simple form by itself. The trick here is to differentiate the `(1-x^-x)` separately, as shown, by equating it to the natural logarithm of any variable `t` and then simplifying the process of differentiation.