Question

What is the derivative of `y=x^(x-1)`?

Answer 1

I got:

`= x^(x-2) (x + xlnx - 1)`

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Assuming you don't remember `d/(dx)[x^x]`, we can still do this via implicit differentiation.

`lny = lnx^(x-1)`

`lny = (x-1)lnx = xlnx - lnx`

Now taking the derivative is a bit easier.

`d/(dx)[lny] = d/(dx)[xlnx - lnx]`

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

`1/x^(x-1) (dy)/(dx) = 1 - 1/x + lnx`

`color(blue)((dy)/(dx)) = x^(x-1)(1 - 1/(x^1) + lnx)`

Finally, we can factor out `1/x` from all terms.

`=> x^(x-1)(x/x*1 - cancel(x)/x*1/cancel(x) + x/x*lnx)`

`= x^(x-2)(x - 1 + xlnx)`

`= color(blue)(x^(x-2)(x + xlnx - 1))`