Let $X_1,...,X_n \overset{iid}{\sim} Uniform(\theta ,2\theta)$, for some positive real $\theta$.
Find the Maximum Likelihood Estimator, $\hat{\theta}$, for $\theta$.
I know the likelihood function is $$L(\theta | X_1,...,X_n) = \prod_{i=1}^n \theta^{-1}1_{\theta \leq X_i \leq 2\theta} = \theta^{-n}1_{\theta \leq X_1,...,X_n \leq 2\theta }.$$
Since the support of $L$ depends on $\theta$, I do not believe differentiating the log-likelihood is helpful. However, since $\theta ^{-n}$ decreases as $\theta$ increases, I would say the maximum of $L$ will occur at $\hat{\theta} $ where $2\hat{\theta}= max X_i =: X_{(n)}$. This means $\hat{\theta} = .5X_{(n)}$ is the maximum likelihood estimator for $\theta$.
Does this seem like a correct idea? If not, any suggestions would be helpful.
All of your observations are correct, the likelihood $L_n$ is given by $$L_n(\theta) = \cases{ \theta^{-n}\,\,\text{ if }X_{(n)}/2\le\theta\le X_{(1)}\\ 0\ \text{ else}}$$ Although $L_n$ is not differentiable on all of $\mathbb R^+$, it suffices to notice that $\theta\mapsto\theta^{-n}$ is a stricly positive and decreasing function on $[X_{(n)}/2,X_{(1)}]$.
It follows from that observation that $L_n(\theta)$ is maximal for the smallest possible value of $\theta$ in $[X_{(n)}/2,X_{(1)}]$, hence $$\hat\theta = X_{(n)}/2 $$ Is the Maximum Likelihood Estimator of $\theta$.