The Hanging Chain Problem, checking if I am right I am trying to make sure I approached this problem correctly. 
It's a hanging chain. I'm told by a classmate that the physics doesn't make sense, but it's a really a math problem, so that I guess is all right – the physics is part of it but not the whole thing. 
We are deriving the PDE that models a hanging chain of Length L. 
The x-axis is placed vertically. Positive direction points upwards. 
The fixed end of the chain (the part attached to the cieling) is at x=L. 
Let $u(x,t)$ denote the deflection (EDIT: the bending) of the chain. We assume the deflection is in the x,u plane. Let $\rho$ denote the mass density in units of mass per length. 
A)  show that in equilibrium position the tension at a point $x$ is $\tau(x)= \rho gx $ where g is the acceleration due to gravity. 
B)  show that $\rho \frac{\partial ^2 u}{\partial t^2} = \frac{1}{ \Delta x}\left[\tau(x+ \Delta x) \frac{\partial u}{\partial x}( x+ \Delta x,t) - \tau(x) \frac{\partial u}{\partial x}(x,t)\right]$
C)  Let $ \Delta x \rightarrow 0$ and obtain $ \rho \frac{\partial ^2 u}{\partial t^2}= \frac{\partial}{\partial x}\left[\tau(x)\frac{\partial u}{\partial x}\right]$
I approached it like this: 
We know F=ma. That means $\tau=ma$ since $a=g$.
The mass of the string at any point x is $\rho x$
Therefore $\tau=\rho g x $
For part B). The component of the tension in the y-axis is going to be  $-\tau \sin \alpha + \tau \sin \beta$ because we are superposing the tension at both ends of the chain. $-\tau \sin \alpha + \tau \sin \beta = ma = \rho g x = \rho g L = \rho g x \Delta x$. Since g is acceleration $ g= \frac{\partial ^2 u}{\partial x^2}$ and that gets us: 
$ -\tau \sin \alpha + \tau \sin \beta =\rho \frac{\partial ^2 u}{\partial x^2} \Delta x$
Now, the interesting thing here is that for most angles $\tau \sin \alpha $ and $ \tau \sin \beta$ will be pretty close to $\tau \tan \alpha $ and $ \tau \tan \beta$. And the tangent on the bent string is going to describe the velocity of the bit of string at that point. 
So we can plug in the tangents to our earlier expression for the tension. 
$ -\tau \tan \alpha + \tau \tan \beta =\rho \frac{\partial ^2 u}{\partial t^2} \Delta x$
Now let's look at $u(x,t)$ in terms of x only. The slope of the tangent to the hanging chain – that's the graph of $u(x,t)$ – is $ \frac{\partial u}{\partial x}(x,t)$ which means 
$ \tan \alpha =  \frac{\partial u}{\partial x}(x,t)$ and $ \tan \beta =  \frac{\partial u}{\partial x}(x+\Delta x,t)$
Plug these back into the equation with the sines of alpha and beta  
$-\tau \tan \alpha + \tau \tan \beta = \left[\rho \frac{\partial ^2 u}{\partial x^2} \Delta x \right] \Rightarrow \tau \left[-\frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t)\right] = \rho \frac{\partial ^2 u}{\partial x^2} \Delta x$
The last step I am a bit fuzzy on. If I move the variables around a bit: 
$$\frac{\left[-\frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t)\right]}{\Delta x} = \frac{\rho}{\tau} \frac{\partial ^2 u}{\partial x^2} $$
I get a zero in the denominator as $\Delta x$ approaches zero. But of course I am assuming I did the rest of the problem right to begin with. I know that 
$\frac{\partial}{\partial x}\left[\tau(x)\frac{\partial u}{\partial x}\right] = \left[\frac{\partial \tau(x)}{\partial x}\frac{\partial u}{\partial x} + \tau(x)\frac{\partial^2 u}{\partial x^2}\right]$
But again I feel like I am just missing the very last and probably dead simple bit at the very end. 
Anyhow sorry to make you all slog through this. 
 A: I came across this question while doing a problem set with a similar exercise, so hopefully this answer is a helpful reference.
You made a few typos after you derived the tangent approximation, where you switched the ordering and sign of the $\beta$ terms, used the $\tau$ for $x$ at $x+\Delta x$, and changed to taking $x$-derivatives on the right hand side.
I fix these below:

Plug these back into the equation with the sines of alpha and beta
$-\tau \tan \alpha + \tau \tan \beta = \left[\rho \frac{\partial ^2 u}{\partial t^2} \Delta x \right] \Rightarrow \left[-\tau(x) \frac{\partial u}{\partial x}(x,t) + \tau(x+\Delta x)\frac{\partial u}{\partial x}(x+\Delta x, t)\right] = \rho \frac{\partial ^2 u}{\partial t^2} \Delta x$

We divide both sides by $\Delta x$:
$$\frac{\left[-\tau(x) \frac{\partial u}{\partial x}(x,t) + \tau(x+\Delta x)\frac{\partial u}{\partial x}(x+\Delta x,t)\right]}{\Delta x} = \rho \frac{\partial ^2 u}{\partial t^2} $$
Now notice that for any function $f$, $$\lim_{\Delta\to 0} \frac{\left[ -f(x) + f(x + \Delta x) \right]}{\Delta x} = \frac{\partial f}{\partial x}(x)$$, by definition of the derivative.
So we take the limit of our last step and substitute this definition in:
$$\frac{\partial\left[ \tau \frac{\partial u}{\partial x}\right]}{\partial x} = \rho \frac{\partial ^2 u}{\partial t^2},$$
which is as required.
