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

Nov 22, 2015

I like the form: $\ln \left(\frac{x + 1}{x}\right) - \frac{1}{x + 1}$

#### Explanation:

Before we differentiate, let's get rid of that silly "divided by $\frac{1}{x}$"

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

We can now use the product rule, but also note that

$1 + \frac{1}{x} = \frac{x + 1}{x}$,

So we have $x \ln \left(\frac{x + 1}{x}\right)$

When we differentiate the second factor, we'll need the chain rule and we'll need $\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x}\right) = \frac{d}{\mathrm{dx}} \left(1 + \frac{1}{x}\right) = 0 - \frac{1}{x} ^ 2 = \frac{- 1}{x} ^ 2$.

So, here we go:

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

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

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

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

Nov 22, 2015

Using the property of logs ...

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

#### Explanation:

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

Using only the product rule :

$y ' = \left(1\right) \left[\ln \left(x + 1\right) - \ln \left(x\right)\right] + x \left[\frac{1}{x + 1} - \frac{1}{x}\right]$

Now simplify ...

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

hope that helped