What is the derivative of $f(x)=ln(x)/x$ ?

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Answer 1 · 2014-09-22T17:29:14

$y'={1/x cdot x-lnx cdot 1}/{x^2}={1-lnx}/{x^2}$

This problem can also be solved by the Product Rule

$y'=f'(x)g(x)+f(x)g(x)$

The original function can also be rewritten using negative exponents.

$f(x)=ln(x)/x=ln(x)*x^-1$

$f'(x)=1/x*x^-1+ln(x)*-1x^-2$

$f'(x)=1/x*1/x+ln(x)*-1/x^2$

$f'(x)=1/x^2-ln(x)/x^2$

$f'(x)=(1-ln(x))/x^2$

What is the derivative of f(x)=ln(x)/xf(x)=ln(x)/x ?