This work is from a textbook where the answer is given as the MLE being $X_{1:n}$ where $1:n$ indicates the minimum value. The pdf of the distribution is given as: $$ f(x;\theta)=\begin{cases}2\theta^2x^{-3}\hspace{10pt}\theta\leq{x}\\0\hspace{33pt}x<\theta;0<\theta\end{cases} $$ Typically, I would calculate the log-likelihood which is $$ \ell(\theta)=n\ln(2)+2n\ln(\theta)-3\sum^n_{i=1}\ln(x_i), $$ and then take the derivative $$ \ell'(\theta)=\frac{2n}{\theta} $$ and then work out the maximum.
Obviously, in this case, it evaluates to zero and I'm not sure what I'm missing to understand how to get to the solution given. In this instance, how should I be calculating the MLE?