\[\begin{align*} \boxed{\dfrac{d}{dx} \ln\dfrac{1}{x} = -\dfrac1x} \end{align*}\]
Solving for the Derivative
Notice that $\ln\dfrac1x = -\ln x$. We can see this if we look at $\ln 1 = 0$. We may rewrite the inside:
\[\begin{align*} \ln 1 &= \ln\left(\dfrac{x}{x}\right)\\ &= \ln\left(x \cdot \dfrac1x\right)\\ \end{align*}\]
We may then apply the product rule for logs to obtain:
\[\begin{align*} &= \ln x + \ln\left(\dfrac1x\right) = 0 \end{align*}\]
For this to be equal to $0$, we must have that $\ln\left(\dfrac1x\right) = -\ln x$. Returning to our original problem, $\dfrac{d}{dx} \ln x = \dfrac{1}{x}$ is a common derivative. Finishing our problem, we take the negative to obtain $\boxed{\dfrac{d}{dx} \ln\dfrac{1}{x} = -\ln x = -\dfrac1x}$
