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
#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
#=> 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))#