Maximum likelihood estimate of $\theta$ with $f(x|\theta) = -(\theta+\theta^2)(1+x)^{\theta-1}x, -10 $ Let $X_1, X_2, ..., X_n$ be a random sample from the pdf $f(x|\theta) =  -(\theta+\theta^2)(1+x)^{\theta-1}x, -1<x<0, \theta>0,$ zero elsewhere. 
How could I deal with the negative sign when I take partial derivative w.r.t $\theta$ ?
 A: Hint: let $A=\prod_{i=1}^n(1+x_i)$ and $B=\prod_{i=1}^n(-x_i)$. You're to find $\theta$ that maximizes
$$
(\theta+\theta^2)A^{\theta-1}B.
$$
Since $\theta$ cannot affect $B>0$, you can just maximize $(\theta+\theta^2)A^{\theta-1}$. Taking derivative:
$$
\partial_\theta((\theta+\theta^2)A^{\theta-1})=A^{\theta-1}[1+\theta(2+\log A)+\theta^2\log A].
$$
When setting the RHS above to $0$, you don't need to worry about $A^{\theta-1}$ so what remains to select the positive root of the quadratic in square brackets above.
A: 
How could I deal with the negative sign...

If you mean the restricion $\theta >0$ : that only gives you the domain of your variable: $\theta$ is restricted to $(0,+\infty)$. To find the (global) maximum of a real function you do as usual: if the function is derivable in all its domain, you find the critical points and check if they are indeed local extremes. Of course, you must only consider the critical points (zeroes of the derivative) that lie inside your domain. That's where the restriction  $\theta >0$ comes into play.
Also (this is important) you must check if your local maxima is really a global maximum (in particular, look at your domain boundary).
