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

1 Answer
Aug 16, 2016

I got:

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


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))#