# What is the derivative of ln(x^2+1)^(1/2)?

Feb 25, 2016

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

#### Explanation:

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

Here we need to apply the chain rule successively:

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

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

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