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

Sep 22, 2014

$y ' = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{{x}^{2}} = \frac{1 - \ln x}{{x}^{2}}$

This problem can also be solved by the Product Rule

$y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g \left(x\right)$

The original function can also be rewritten using negative exponents.

$f \left(x\right) = \ln \frac{x}{x} = \ln \left(x\right) \cdot {x}^{-} 1$

$f ' \left(x\right) = \frac{1}{x} \cdot {x}^{-} 1 + \ln \left(x\right) \cdot - 1 {x}^{-} 2$

$f ' \left(x\right) = \frac{1}{x} \cdot \frac{1}{x} + \ln \left(x\right) \cdot - \frac{1}{x} ^ 2$

$f ' \left(x\right) = \frac{1}{x} ^ 2 - \ln \frac{x}{x} ^ 2$

$f ' \left(x\right) = \frac{1 - \ln \left(x\right)}{x} ^ 2$