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.

$f(x) = e^{\ln(1/x)}=1/x$, then $f'(x) =(1/x)'=\frac{1'\cdot x-x'\cdot 1}{x^2}=\frac{- 1}{x^2}$
