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.

Hint: the exponential and logarithm are inverse functions:
$$e^{\ln u}=u$$
for any $u>0$.
