What is the derivative of $y＝lnx/ x$ ?

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Answer 1 · 2016-01-29T04:55:47

$y'=(1-lnx)/x^2$

Use the quotient rule, which states that

$d/dx[(f(x))/(g(x))]=(f'(x)g(x)-g'(x)f(x))/[g(x)]^2$

Applying this to $y=lnx/x$ , we see that

$y'=(xd/dx[lnx]-lnxd/dx[x])/x^2$

Since $d/dx[lnx]=1/x$ and $d/dx[x]=1$ ,

$y'=(x(1/x)-lnx)/x^2$

$y'=(1-lnx)/x^2$

What is the derivative of y＝lnx/ xy＝lnx/ x?