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Having a function $f(t)$, what is $\dfrac{\mathrm d(\mathrm df/\mathrm dt)}{\mathrm df}$ equal to?

I attempted to approach this using the first principles (of differentiation) and got a limit of the form $\frac00$. Is the answer to this simply $0$? And, if so, why?

grg
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Darbininkai Broliai
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  • Do you really mean for your second differentiation to be "with respect to $f$"? (It's not clear what that means; this post may be of interest.) – Andrew D. Hwang Mar 08 '15 at 14:44
  • The question comes from trying to solve the double pendulum using the lagrange equation. I have some terms of the form dtheta/dt and I need to differentiate these terms with respect to theta, where theta is some function of time that I dont know yet. – Darbininkai Broliai Mar 08 '15 at 14:48
  • It sounds as if you may be using $\theta$ to stand for two distinct things: A coordinate on your configuration space, and an unknown function of time (whose value at time $t$ is a configuration space coordinate of the system). If that's right, it's "safer" to use a different letter for the unknown function, perhaps $\Theta$. Then your question becomes, "What is $d(d\Theta/dt)/d\theta$?" I hope that helps, but if not, perhaps post more details of your calculation...? – Andrew D. Hwang Mar 08 '15 at 15:04
  • It is indeed a coordinate. But i think it is simply not a function, as taking the answer to be 0 gives me the correct result in the end – Darbininkai Broliai Mar 08 '15 at 15:12

1 Answers1

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The "stenographic" derivation of the desired formula proceeds as follows: You have the three variables $t$, then $y:=f(t)$, and finally $v:=f'(t)$, and you want to know ${dv\over dy}$. The chain rule gives $${dv\over dy}={dv/dt\over dy/dt}={f''(t)\over f'(t)}\ ,\tag{1}$$ and that's it.

The above looks like magic. So let's do it once over more carefully. You are given a function $t\mapsto y:=f(t)$ in some $t$-interval $I$. Assume that $f'(t)\ne0$ on $I$. Then $f$ maps $I$ bijectively onto some $y$-interval $J$, and has an inverse function $$g:=f^{-1}:\quad J\to I,\qquad y\mapsto t=g(y)\ .$$ You are interested in the velocity $v(t):=f'(t)$as a function of $y$, i.e., in the function $$\hat v(y):=f'\bigl(g(y)\bigr)\ .$$ According to the chain rule the derivative of $\hat v$ computes to $$\hat v'(y)=f''\bigl(g(y)\bigr)\>g'(y)=f''\bigl(g(y)\bigr)\>{1\over f'\bigl(g(y)\bigr)}\ ,$$ and this is formula $(1)$ expressed in terms of the variable $y$.

Christian Blatter
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