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

Mar 4, 2018

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sqrt{{x}^{2} + 1}\right)\right) = \frac{x}{{x}^{2} + 1}$

#### Explanation:

Using the chain rule:

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sqrt{{x}^{2} + 1}\right)\right) = \frac{1}{\sqrt{{x}^{2} + 1}} \frac{d}{\mathrm{dx}} \sqrt{{x}^{2} + 1}$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sqrt{{x}^{2} + 1}\right)\right) = \frac{1}{\sqrt{{x}^{2} + 1}} \frac{1}{2 \sqrt{{x}^{2} + 1}} \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sqrt{{x}^{2} + 1}\right)\right) = \frac{1}{2 \left({x}^{2} + 1\right)} \left(2 x\right)$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sqrt{{x}^{2} + 1}\right)\right) = \frac{x}{{x}^{2} + 1}$

We can also note that using the properties of logarithms:

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

and simplify the passages:

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sqrt{{x}^{2} + 1}\right)\right) = \frac{1}{2} \frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 1\right) = \frac{1}{2} \frac{1}{{x}^{2} + 1} \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = \left(\frac{1}{2}\right) \left(\frac{1}{{x}^{2} + 1}\right) \left(2 x\right) = \frac{x}{{x}^{2} + 1}$