Question

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

Answer 1

`f'(x) = -(cos((1)/(ln x)))/(xln^2x)`

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(u) = sin(u)`

where `u(t) = (1)/(t)`

and `t(x) = ln x`

So, the derivative of `f(x)` is (using the chain rule):

`f'(x) = (df)/(du)(du)/(dt)(dt)/(dx)`

We calculate these derivatives:

`(df)/(du) = cos(u)`

`(du)/(dt) = -(1)/(t^2)`

`(dt)/(dx) = (1)/(x)`

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

`f'(x) = cos(u)(-1)/(t^2)(1)/(x) = -cos((1)/(lnx))(1)/(xln^2x)`