Statement: If f, g are continuous on [a,b], differentiable on (a,b), and g'(x) != 0 on (a,b), then there exists c in (a,b) with f'$\frac{c}{g}$'(c) = [f(b)-f(a)]/[g(b)-g(a)].

Special case: When g(x) = x, this reduces to LMVT (since g'(c) = 1 and g(b)-g(a) = b-a).

Connection to L'Hopital's Rule: L'Hopital's Rule can be proved using Cauchy's MVT. For the 0/0 form: lim f$\frac{x}{g}$(x) = lim f'$\frac{c}{g}$'(c) as the interval shrinks to a point.

Why g'(x) != 0 is needed: Ensures g(b) != g(a) (so the ratio makes sense) and prevents division by zero in f'$\frac{c}{g}$'(c).