# What are the first and second derivatives of f(x)=(ln x)^(1/2)?

Apr 15, 2016

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

and

$f ' ' \left(x\right) = - \frac{1}{2 {x}^{2}} \left[\frac{1}{2 \sqrt{{\ln}^{3} x}} + \frac{1}{\sqrt{\ln x}}\right]$

#### Explanation:

To get the first derivative, we'll use the chain rule :

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

the second derivative again uses first the product rule and then the chain rule:

$f ' ' \left(x\right) = \frac{d f ' \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \cdot \frac{1}{x} = - \frac{1}{4} {\left(\ln x\right)}^{- \frac{3}{2}} \frac{d}{\mathrm{dx}} \left(\ln x\right) \cdot \frac{1}{x} + \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \left(- \frac{1}{x} ^ 2\right) = - \frac{1}{4} {\left(\ln x\right)}^{- \frac{3}{2}} \frac{1}{x} ^ 2 - \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \frac{1}{x} ^ 2$

We can also choose to write these two answers in simpler forms:

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

and

$f ' ' \left(x\right) = - \frac{1}{2 {x}^{2}} \left[\frac{1}{2 \sqrt{{\ln}^{3} x}} + \frac{1}{\sqrt{\ln x}}\right]$