# Differentiate f(x) = lnabs(x/(1+x^2))?

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Rhys Share
Jan 25, 2018

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

#### Explanation:

We must use our log laws:

${\log}_{\alpha} \left(\frac{\beta}{\gamma}\right) \equiv {\log}_{\alpha} \beta - {\log}_{\alpha} \gamma$

$\implies \ln | \frac{x}{1 + {x}^{2}} | = \ln | x | - \ln | 1 + {x}^{2} |$

Now applying our knowledge of differentiating logs:

$\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$

By using the chain rule...

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

color(blue)(= 1/x - (2x)/(1+x^2)

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