How do I partially differentiate an equation? I've so far only differentiated functions (for eg. $ f(x,y) = x^2 + Y^2$). How do I differentiate equations? For example, here's a constraint equation that I'm trying to partially differentiate with respect to x:
$$ P_x X + P_y Y = B $$
(B is a constant; $ P_x $ and $ P_y $ are constants) The answer that my reference material is giving me is: $P_x$
If we partially differentiate both sides with respect x, we'll get 1 = 0, right? What is my reference material trying to mean when it says that the answer is 1?
 A: It is not 100% clear what you are trying to do, so I'll answer as best as I can.  Let's start by finding the total differential.  In your case, since
$$P_x x + P_y y = B$$
then the total differential will be
$$P_x\,dx + P_y\,dy = 0 $$
Therefore, $\frac{dy}{dx}$ will be
$$\frac{dy}{dx} = -\frac{P_x}{P_y}.$$
So, in this case, there are only two differentials.  Since one is independent and the other is dependent, then the partial derivative is the same as the total derivative.  In other words (even though I hate this notation),
$$\frac{\partial y}{\partial x} = -\frac{P_x}{P_y}.$$
However, sometimes the question is to rewrite the whole thing as a function of $x$ and $y$, such that $F(x, y) = 0$, and the goal is to find $\frac{\partial F}{\partial x}$.
In that case, you would start by rewriting the equation as
$$F(x, y) = P_x x + P_y y - B = 0.$$
Then, the total differential would be
$$dF = P_x\,dx + P_y\,dy = 0$$
The partial differential is found by setting all the other independent differentials (i.e., $dy$) to zero:
$$\partial F = P_x\,\partial x + 0$$
$$\partial F = P_x\,\partial x$$
$$\frac{\partial F}{\partial x} = P_x$$
By the way, the notation I prefer to use for partials is this, as it clearly demarcates what the components of the fraction are:
$$\frac{\partial_x F}{dx} = P_x$$
However, that notation is rarely, if ever, used.  It says that we are looking at the partial differential of $F$ that occurs when $x$ is allowed to vary, divided by the infinitesimal change in $x$, which is allowed to freely vary.
Also, because $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} = 0$, then $\frac{\partial F}{\partial y} = -\frac{\partial F}{\partial x}$.
