How do you take the derivative of a function's derivative ...

[Sandstorm]

I know perfectly how normal derivatives work, like d/dx f(x) = f'(x). But how would you evaluate d/dV V'(t)? It's obvious that d/dt V'(t) would be V"(t) or the 2nd derivative of V. What about d/dV?

[Jones]

f(x) = x^2
f'(x) = 2x
f''(x) = 2

[aero]

Suppose you know

V = f(t)
V' = f'(t)

You want to find d/dV (V').  This means "As V is changed slightly, what is the resultant change in V'.
The only variable linking them together is t.

dV/dt = f'(t)
dV'/dt = f''(t)

So dt/dV = 1/f'(t) by the inverse function theorem.  The key thing you need to do this is that V doesn't have derivative 0.

Therefore dV'/dV = dV'/dt * dt/dV = f''(t) / f'(t).