# What is the derivative of f(x)=sin(1/lnx)?

Jan 26, 2016

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

#### Explanation:

We need to use the chain rule.

First, we can rename the function to make the calculations easier. So, we have a few functions called:

$f \left(u\right) = \sin \left(u\right)$

where u(t) = (1)/(t)

and $t \left(x\right) = \ln x$

So, the derivative of $f \left(x\right)$ is (using the chain rule):

$f ' \left(x\right) = \frac{\mathrm{df}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

We calculate these derivatives:

$\frac{\mathrm{df}}{\mathrm{du}} = \cos \left(u\right)$

$\frac{\mathrm{du}}{\mathrm{dt}} = - \frac{1}{{t}^{2}}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{1}{x}$

The final expression is the multiplication of them and the substitution of their previous definitions:

$f ' \left(x\right) = \cos \left(u\right) \frac{- 1}{{t}^{2}} \frac{1}{x} = - \cos \left(\frac{1}{\ln x}\right) \frac{1}{x {\ln}^{2} x}$