What is the derivative of #lnx/x^2#?

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Answer 1 · 2016-12-03T10:31:32

#f'(x)=(1-lnx^2)/x^3#

use the quotient rule, which should be memorised.

#f(x)=u/v" "=>" "f'(x)=(vu'-uv')/v^2#

#f(x)=(lnx)/x^2#

#u=lnx=>u'=1/x#

#v=x^2=>v'=2x#

#f'(x)=(x^2xx1/x-2xlnx)/(x^2)^2#

#f'(x)=(x-2xlnx)/x^4=(x(1-2lnx))/x^4=(cancel(x)(1-lnx^2))/(cancel(x^4)x^3)#