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

Aug 30, 2015

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

Explanation:

Although you could use $\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$, the algebra will get messy that way.

Let's rewrite using properties of $\ln$.

$y = \ln \left(1 + \left(\frac{1}{x}\right)\right) = \ln \left(\frac{x + 1}{x}\right)$

$= \ln \left(x + 1\right) - \ln \left(x\right)$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x + 1} - \frac{1}{x} = \frac{- 1}{x \left(x + 1\right)}$