Use logarithmic differentiation to find the derivative of the function y = x^x

Question

anonymous · Answer

(1+logx)x^x

lgbasallote · Answer

y = x^x

ln both sides...

lny = lnx^x

via reverse power rule...

lny = xlnx

you can use implicit differentiation now...you do know that right/

anonymous · Answer

so d/dx(lny) = 1/y * y' and d/dx(xlnx) = x*1/x .....????

anonymous · Answer

it goes lyk tis  logy= xlogx
diff wrt to x on bth sides so on lhs its 1/y dy/dx=1+logx by product rule on rhs  u get 1+logx 
thn dy/dx =(1+logx)y wer y is nthin bt =x^x

anonymous · Answer

one of my favourite derivatives. an alternative to the ones suggested above:
y = e^log(x^x) = e^xlog(x) which you can differentiate using the chain rule

anonymous · Answer

thx guys you just walked me through my first ever log diff problem