What are the critical values, if any, of #f(x)= x^6ln(x) #?

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Answer 1 · 2018-07-04T16:15:43

#x=0# or #x=e^(-1/6)#

#(uv)'=u'v+uv'#
and the fact that

#(ln(x))'=1/x#

so we get

#f'(x)=6x^5+x^6*1/x#

#f'(x)=x^5(6ln(x)+1)#
solving #f'(x)=0#
we get

#x=0# or #x=e^(-1/6)#