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

Aug 25, 2016

(x^2-1)/(x(x^2+1)

#### Explanation:

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

$f ' \left(x\right) = \frac{1}{x + \frac{1}{x}} \cdot \frac{d}{\mathrm{dx}} \left(x + \frac{1}{x}\right)$ (Standard differential and Chain rule)

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

$= \frac{x}{{x}^{2} + 1} \cdot \frac{{x}^{2} - 1}{x} ^ 2$

$= \frac{1}{{x}^{2} + 1} \cdot \frac{{x}^{2} - 1}{x}$

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

Aug 25, 2016

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

#### Explanation:

$y = \ln \left(x + \frac{1}{x}\right)$
Let $\left(x + \frac{1}{x}\right) = z$
so $y = \ln z$
$\frac{\mathrm{dy}}{\mathrm{dz}} = \frac{1}{z}$

$z = x + {x}^{-} 1$
$\frac{\mathrm{dz}}{\mathrm{dx}} = 1 - {x}^{-} 2$

We are using the chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dz}} \cdot \frac{\mathrm{dz}}{\mathrm{dx}}$
And $\left(x + \frac{1}{x}\right) = \frac{{x}^{2} + 1}{x}$
And $\left(1 - {x}^{-} 2\right) = \frac{{x}^{2} - 1}{x} ^ 2$

So$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{{x}^{2} + 1} \cdot \frac{{x}^{2} - 1}{x} ^ 2$

And we have the answer