Let $X_1,\dots,X_n$ be independent random samples from distribution with density $$f(x) = \frac{\theta e^{\theta(x-\mu)}}{(1+e^{\theta(x-\mu)})^2}$$ for $x$ real, and $\mu$ from $\Bbb R$, $\theta >0$. We are asked to find MLE of $\mu$ and $\theta$.
I tried to obtain the joint density, and took the log-likelihood. But I cannot deal with this denominator. Any hint will be appreciated! Thanks!
So Joint likelihood is $\theta^n e^{-\mu \theta n} \frac{e^{\theta \sum x_i}}{(1 + e^{\theta(x_1 - \mu)})^2 ... (1 + e^{\theta(x_n - \mu)})^2}$
Taking log, we have:
$\log \theta^n - \mu \theta n + \theta \sum x_i - 2 \sum \log(1 + e^{\theta (x_i - \mu)})$
I believe taking partial derivatives with respect $\mu$ and $\theta$ can get you the results.