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

Jul 16, 2016

Simplify and apply the chain rule to find that

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

#### Explanation:

To make this a little easier, first we will simplify the expression to

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

Now, using the product rule, chain rule, and the derivatives $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$ and $\frac{d}{\mathrm{dx}} \frac{1}{x} = - \frac{1}{x} ^ 2$, we have

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

(simplification)

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

(product rule)

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

(chain rule and derivatives of $\ln \left(x\right)$ and $x$)

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

(derivative of $\frac{1}{x}$)

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

(simplification)