Question

What is the derivative of `y＝lnx/ x`?

Answer 1

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

Explanation:

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`