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.
 A: Hint:
$$\quad e^{\ln(a)}=a\quad $$
Solving the more complicated way also gives the right answer:
$$\frac{d}{dx}(e^{\ln(1/x)})=e^{\ln(1/x)}\cdot\frac{d}{dx}(\ln(1/x))=e^{\ln(1/x)}\cdot \left[\frac{1}{1/x}\cdot \frac{d}{dx}(1/x)\right]=\frac{1}{x}\cdot \left[x\cdot \left(-\frac{1}{x^2}\right)\right]$$
where we have used that
$$\frac{d}{dt}(e^{f(t)})=e^{f(t)}\left(\frac{d}{dt} f(t)\right)$$
and that
$$\frac{d}{dt}(\ln(g(t)))=\frac{1}{g(t)}\left(\frac{d}{dt} g(t)\right)$$
A: Hint: the exponential and logarithm are inverse functions:
$$e^{\ln u}=u$$
for any $u>0$.
A: 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.
A: $f(x) = e^{\ln(1/x)}=1/x$, then $f'(x) =(1/x)'=\frac{1'\cdot x-x'\cdot 1}{x^2}=\frac{- 1}{x^2}$
A: The other hints provided will help you find the correct solution the easy way.
However, I'd like to point out that your approach has a flaw:
$$\frac{d}{dx}\left(e^{f(x)}\right)\ne e^{f(x)}\cdot f(x)$$
Rather, it is:
$$\frac{d}{dx}\left(e^{f(x)}\right)= e^{f(x)}\cdot f'(x)$$
Thus, we'd have (this is the hard way, applying the line of thought you tried to use):
$$\begin{align}
\frac{d}{dx}\left(e^{\ln 1/x}\right) & = e^{\ln 1/x}\cdot\left(\frac{d}{dx}\left(\ln 1/x\right)\right)\\
& = e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\frac{d}{dx}\left(1/x\right)\right)\\
& = e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right)\\
& = e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right)\\
\end{align}$$
This is the correct answer, we don't simplify anything.  Simplifying (using the same property as other have hinted that you should use:
$$\begin{align}
e^{\ln 1/x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right) & = \frac{1}{x}\cdot\left(\frac{1}{1/x}\cdot\left(\frac{-1}{x^2}\right)\right)\\
 & = \frac{1}{x}\cdot\left(x\cdot\left(\frac{-1}{x^2}\right)\right)\\
 & = \frac{-1}{x^2}
\end{align}$$
A: As mentioned by others you use that 
$$
e^{\ln(x)} = x
$$
so that 
$$
f(x) = \frac{1}{x}.
$$
Note here though that the domain of $f$ is all positive numbers. The original expression for $f$ is not defined for negative $f$. So also the domain for the derivative will (only) be all positive numbers.
A: Applying the chain rule, one gets
$$
\frac{d}{dx} e^{\ln(1/x)} = e^{\ln(1/x)}\cdot \ln'(1/x)\cdot \frac{d}{dx} \frac1x =\cdots\cdots
$$
So:
$$
\frac{d}{dx} e^{\ln(1/x)} = e^{\ln(1/x)}\cdot \frac{1}{1/x}\cdot \frac{d}{dx} \frac1x =\cdots\cdots
$$
You were simply failing to apply the chain rule correctly.
All of the above is of course the hard way.  Here's the easy way:
$$
\frac{d}{dx} e^{\ln(\text{whatever})} = \frac{d}{dx}(\text{whatever}) = \cdots\cdots
$$
