Can you differentiate x^x?

2 Answers
Jan 3, 2018

#=> (dy)/(dx) = x^x ( lnx +1) #

Explanation:

This question will require the knowledge of implicit differentiation and logarithms:

Let: # y = x^x#

Taking natural logs:

# lny = lnx^x #

Now using our knowledge of logs :

# log_gamma alpha^beta -= beta log_gamma alpha #

#=> lny = xlnx #

Implicitly differentiating, also using product rule

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

As #color(blue)(d/(dx) ( lnx ) = 1/x #

and #color(green)(d/(dx) xlnx = x* d/(dx) ( lnx) + lnx * d/(dx) ( x ) #

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

#=> color(red)((dy)/(dx) = x^x ( lnx +1) #

Graphs:

# color(purple)(y= x^x #:
desmos

#color(orange)( y = d/(dx) ( x^x ) = x^x ( lnx + 1 ): #

desmos

Jan 4, 2018

#d/dx (x^x) = x^x(1+lnx)#

Explanation:

You can also differentiate explicitly by writing the function as:

#x^x = (e^lnx)^x = e^(xlnx)#

so that:

#d/dx (x^x) = d/dx (e^(xlnx)) #

and then using the chain rule:

#d/dx (x^x) = e^(xlnx) d/dx(xlnx) = x^x(1+lnx)#