How do you find the derivative of #y=x^2+x^x#?

1 Answer
Jul 17, 2015

We know the first half. Let's try the second half, using a more foolproof way than simply trying to remember the derivative of a similar function.

Let:
#y = x^x#

#log_x(y) = log_x(x^(x)) = x#

#(lny)/(lnx) = x#

#lny = xlnx#

#1/y ((dy)/(dx)) = x*1/x + lnx*1#
(Implicit Differentiation, Product Rule)

#1/y ((dy)/(dx)) = 1 + lnx#

#((dy)/(dx))_(y = x^x) = y[1 + lnx]#

#= x^x[1 + lnx]#

Thus you have overall:

#color(blue)(((dy)/(dx))_(y = x^2 + x^x) = 2x + x^x(1+lnx))#