# How do you find the derivative of ln(x/(x^2+1))?

Jan 22, 2017

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

#### Explanation:

The easiest way is to first rewrite the function using properties of logarithms. Recall that $\log \left(\frac{a}{b}\right) = \log \left(a\right) - \log \left(b\right)$.

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

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

Now we have two simpler functions to differentiate. Recall that $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$. The chain rule tells us that the derivative of a function within $\ln \left(x\right)$ is $\frac{d}{\mathrm{dx}} \ln \left(g \left(x\right)\right) = \frac{1}{g \left(x\right)} \cdot g ' \left(x\right) = \frac{g ' \left(x\right)}{g \left(x\right)}$.

Then:

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

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

These can be combined, but it's not necessary:

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

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