As ever, the first stage to solving this problem is to draw a diagram that depicts the situation, such as the diagram shown below.  This shows a section of the hanging chain. 

We begin by noting that the chain is simply hanging under its own weight – it isn’t moving or accelerating, and so the forces acting on any section of the chain are in balance.  Consider a section AB of the chain, as shown in the diagram above.  There are three forces acting on section AB: 

1.         A horizontal tension force T_H acting at A; 

2.         A tangential tension force T acting at B; 

3.         The vertical weight of the section AB. 

The horizontal and vertical components of these forces must balance (otherwise the hanging chain will not remain stationary).  This gives 

where  is the mass per unit length of the chain, and s is the arc length.  For simplicity, we will consider the case where the mass per unit length is a constant.  Hence 

From the geometry in the blue inset in the diagram above, and the first of the force balance equations, we have: 

and hence by using the second force balance equation and differentiating both sides with respect to x, we get 

Now we make the following variable change: 

Substituting into the previous equation and rearranging, the solution of the problem becomes one of evaluating the integral 

where we have noted that by symmetry if the bottom of the chain is located at x = 0 (as in the diagram above) then the gradient p at that point is also zero. 

This integration is easily carried out to give 

and hence (neglecting the constant of integration)

This curve is known as a catenary.  An example is shown in the following diagram, with the parameter a set to unity.