Derivative of $e^{\ln(1/x)}$ This question looks so simple, yet it confused me.

If $f(x) = e^{\ln(1/x)}$, then $f'(x) =$ ?

I got $e^{\ln(1/x)} \cdot \ln(1/x) \cdot (-1/x^2)$.
And the correct answer is just the plain $-1/x^2$. But I don't know how I can cancel out the other two function.

For $a>0$, $e^{\ln(a)} = a.\;\;$  Similarly, $\;\;\ln(e^a) = a\ln(e) = a\;\;$ because for $$\;f(x) = e^x,\;\;f^{-1}(x) = \ln(x)\quad\text{and}\quad g(x) = \ln x,\;\;g^{-1}(x) = e^x$$
That is, $\ln(x)$ and $e^x$ are each others' inverse function on the set of positive real numbers.
