derivative of log and reciprocal

Question

I am searching for a function $f(x)$ with the following property

The derivative $$\frac{d}{dx}(\frac{1}{f(x)})$$ is in opposite sign with the $$\frac{d}{dx}(ln(\frac{1}{f(x)}))$$ i-e either of two is positive or negative .

Answer 1

$$\left(\log\frac1{f(x)}\right)'=\frac{\left(\dfrac1{f(x)}\right)'}{\dfrac1{f(x)}}$$ so that to achieve opposite signs you need ${f(x)}<0$. But then the logarithm is undefined.

Answer 2

This looks like a strange problem... I assume you are looking for a differentiable function f : \mathbb{R} \rightarrow \mathbb{R}^+, to have a well-defined ln(1/f). But then,

(ln \circ 1/f)’=\frac{1/f}{(1/f)’},

which has the same sign as (1/f)’since ftakes positive values.

Answer 3

$$\frac d{dx}\left(\frac1{f(x)}\right)=-\frac{f'(x)}{f(x)^2}$$ $$\frac d{dx}\left(\ln\left(\frac1{f(x)}\right)\right)=-\frac{f'(x)}{f(x)}$$ So, $$\frac d{dx}\left(\frac1{f(x)}\right)=-\frac d{dx}\left(\ln\left(\frac1{f(x)}\right)\right)\implies -\frac{f'(x)}{f(x)^2}=\frac{f'(x)}{f(x)}$$ If this has to be true for any $x$, I do not see anything to be possible.