How does a parameter affect roots of an equation Assume I have a function that I want to find its roots, lets say $f(x,\theta) = 0$. $x$ is our variable and $\theta$ is a parameter. I know that $\frac{\partial f}{\partial x} \ge 0$ and $\frac{\partial f}{\partial \theta} \le 0$ always holds true. Unfortunately the equation $f(x,\theta) = 0$ can not be solved directly and get a closed form solution for $x$, but I need to study the effects of changing $\theta$ on the root of $f(x,\theta)$. The domain of the function $f$ is of the form $x\ge \alpha \theta$ where $\alpha$ is a known positive constant. Also $\theta$ and $x$ both are positive real variables.
My question is, given the fact that we know $f(x,\theta)$ only has a single unique root, how does increasing or decreasing $\theta$ affects the location of the root?
Thanks in advance.
 A: Do you know the implicit function theorem? It deals with this situation, if you know that $ \frac{\partial f}{\partial x}\neq 0$. Even if this is not true you can still do a formal calculation:
Since you say that $f(x, \theta) =0$ has a single root $x_0$ (given $\theta$), we can formally say $x_0= x_0(\theta)$ and will have
$$f(x_0(\theta) ,\theta) =0$$
Differentiating this implies
$$\frac{\partial f}{\partial x} \frac{dx_0}{d\theta} = - \frac{\partial f}{\partial \theta}$$ -- here I'm already assuming that $x_0$ depends differentiably on $\theta$, which may not be true if $ \frac{\partial f}{\partial x}$ has a zero, but is, otherwise, a conclusion of the implicit function theorem. From this you may already deduce some information on the derivative of $x_0$ wr to $\theta$.
If you know that $ \frac{\partial f}{\partial x}\neq 0$ along the solution you can now get
$$\frac{dx_0}{d\theta}  = - \frac{\frac{\partial f}{\partial \theta}}{\frac{\partial f}{\partial x}} $$
If you now know the both derivatives in the fraction, you have the rate of change of $x_0$ as a function of $\theta$.
