We have:
#y=x^x# Let's take the natural log on both sides.
#ln(y)=ln(x^x)# Using the fact that #log_a(b^c)=clog_a(b)#,
#=>ln(y)=xln(x)# Apply #d/dx# on both sides.
#=>d/dx(ln(y))=d/dx(xln(x))#
The chain rule:
If #f(x)=g(h(x))#, then #f'(x)=g'(h(x))*h'(x)#
Power rule:
#d/dx(x^n)=nx^(n-1)# if #n# is a constant.
Also, #d/dx(lnx)=1/x#
Lastly, the product rule:
If #f(x)=g(x)*h(x)#, then #f'(x)=g'(x)*h(x)+g(x)*h'(x)#
We have:
#=>dy/dx*1/y=d/dx(x)*ln(x)+x*d/dx(ln(x))#
#=>dy/dx*1/y=1*ln(x)+x*1/x#
#=>dy/dx*1/y=ln(x)+cancelx*1/cancelx#
(Don't worry about when #x=0#, because #ln(0)# is undefined)
#=>dy/dx*1/y=ln(x)+1#
#=>dy/dx=y(ln(x)+1)#
Now, since #y=x^x# , we can substitute #y#.
#=>dy/dx=x^x(ln(x)+1)#
