Question

How do you differentiate `y=x^(x-1)`?

Answer 1

Use logarithmic differentiation to get `y'=(lnx+1-1/x)x^(x-1)`.

Explanation:

Take the natural logarithm of both sides, to drop the exponent:
`lny=(x-1)lnx`

Now differentiate both sides with respect to `x`:
`d/dx(lny=(x-1)lnx)`

Note that this will require knowledge of implicit differentiation, because the derivative of `lny` w.r.t.x is `1/y*y'`. The derivative of `(x-1)lnx` is found with the product rule:
`d/dx((x-1)(lnx))=(x-1)'(lnx)+(x-1)(lnx)'`
`=(1)(lnx)+(x-1)*(1/x)`
`=lnx+(x-1)/x`
`=lnx+1-1/x`

Since `d/dx(lny)=1/y*y'`, and `d/dx((x-1)(lnx))=lnx+1-1/x`, we have:
`1/y*y'=lnx+1-1/x`

Multiply both sides by `y` to isolate `y'`:
`y'=(lnx+1-1/x)y`

Since `y=x^(x-1)`:
`y'=(lnx+1-1/x)x^(x-1)`