What is the derivative of $\frac{ln x}{x^{2}}$ $lnx/x^2$ ?

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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.

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

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

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

$v=x^2=>v'=2x$

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

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

What is the derivative of lnx/x^2lnxx2lnx/x^2?