# In(x/x^2+1) diffrentiate?

Jan 18, 2018

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

#### Explanation:

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

For $f$ to be defined in $\mathbb{R}$ we need $\frac{x}{{x}^{2} + 1} > 0$

${x}^{2} + 1$ is always $> 0$ so we need $x > 0$

As a result Domain is ${D}_{f} = \left(0 , + \infty\right)$

For $x$$\in$${D}_{f}$:

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

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

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

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

$- \left(\frac{\cancel{{x}^{2} + 1}}{x}\right) \left(\frac{{x}^{2} - 1}{{x}^{2} + 1} ^ \cancel{2}\right)$ $=$

$- \frac{{x}^{2} - 1}{x \left({x}^{2} + 1\right)}$ $=$ $- \frac{{x}^{2} - 1}{{x}^{3} + x}$