I'm confused on how I am supposed to differentiate the following using the chain rule.

I have $x(t)$ and I have defined new time variables $t=\tau$ and $T=\epsilon t$

I have that

$$ \frac{dx}{dt} = \frac{d \tau}{dt} \frac {\partial x}{\partial \tau} + \frac {dT}{dt} \frac {\partial x}{\partial T} $$

which becomes

$$ \frac {dx}{dy} = \frac {\partial x}{\partial \tau} + \epsilon \frac {\partial x}{\partial T} $$

Then I want to differentiate with respect to $t$, so

$$ \frac {d^2x}{dt^2}= \frac {d}{dt} \left(\frac {\partial x}{\partial \tau}\right)+\epsilon \frac {d}{dt}\left(\frac {\partial x}{\partial T}\right)$$

but I am unsure on how to compute the last part with workings. I have the answer to it but I am unsure on what the steps are and how to differentiate the parital derivatives.

Thanks in advance



We have ${d \over dt}{\partial x \over \partial \tau} + \varepsilon{d \over dt}{\partial x \over \partial T} = {\partial \over \partial \tau}\big({\partial x \over \partial \tau}\big) {d \tau \over dt} + {\partial \over \partial T}\big({\partial x \over \partial \tau}\big) {dT \over dt} + \varepsilon{\partial \over \partial \tau}\big({\partial x \over \partial T}\big) {d \tau \over dt} +\varepsilon{\partial \over \partial T}\big({\partial x \over \partial T}\big) {d T \over dt}$ which gives the desired result when simplified as in the OP.

  • $\begingroup$ I'm supposed to get $\frac {\partial ^2 x}{\partial \tau ^2} +2 \epsilon \frac {\partial ^2 x}{\partial \tau \partial T} +\epsilon ^2 \frac {\partial ^2 x}{\partial T^2}$. and my main issue is how you split up the partial derivatives and differentiate them. $\endgroup$ – juper Apr 16 '18 at 13:08
  • $\begingroup$ Right, I've fixed it now. $\endgroup$ – Lukas Kofler Apr 16 '18 at 13:20
  • $\begingroup$ Ah yes, have figured out the steps to do this now. Thanks for your help! $\endgroup$ – juper Apr 16 '18 at 13:24
  • $\begingroup$ The fact that ${\partial x \over \partial \tau}$ is the derivative of another function doesn't matter when you differentiate it. To see this, you could simply put $y(t) ={\partial x \over \partial \tau}$ and proceed as usual when differentiating with respect to $t$. $\endgroup$ – Lukas Kofler Apr 16 '18 at 13:27
  • 1
    $\begingroup$ yes, this makes perfect sense. Thanks! $\endgroup$ – juper Apr 16 '18 at 13:28

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