>>13705193It's a differential.
The function s is defined to be the integral of that square root, therefore it's derivative is the square root itself.
ds is the differential of s. It is usually defined like this:
You know that ds/dx = s' where s' is the derivative of s and thus it is the square root of 1 + [f'(x)]^2.
The differential of s is then defined to be s'dx, so that if you were to treat the derivative ds/dx as if it were a quotient by the relation ds = s'dx you could get ds/dx = s' by diving through by dx. Derivatives aren't actually quotients (but rather limits of quotients) so this is just a mnemonic device.
You can just forget about the above discussion and just define the differential of x as the difference between the two values of x at the endpoints of an interval. So if the interval is [a, b] the differential dx is defined to be: dx = b - a. In other words, for the identity fuction f(x) = x the differential is simply what is usually dentoed as ?x:
dx = ?x = b - a.
For functions other than the identity function f(x) = x define the differential to be the product of the derivative of f and dx (which is equal to ?x):
df = f'dx = f'?x = f'(b - a).
The point of the differential is that it is a good linear approximation of the behavior of the function when [a, b] is a small interval.
