# Changing denominator in second derivatives?

If $\frac{d^{2}x}{d\tau^{2}}=k\left(\frac{dt}{d\tau}\right)^{2}$ and I then multiply both sides by $\left(\frac{d\tau}{dt}\right)^{2}$ I get $\frac{d^{2}x}{dt^{2}}=k$ . Why does the $d\tau^{2}$ on the left hand side change to $dt^{2}$? $\frac{d^{2}x}{d\tau^{2}}$ is a second derivative, so surely I can't just cancel the $d\tau's$ as in a normal fraction?

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This only works if $\frac{\mathrm d t}{\mathrm d\tau}$ is constant. Otherwise
$$\begin{eqnarray} \frac{\mathrm d^2x}{\mathrm d\tau^2} &=&\frac{\mathrm d}{\mathrm d\tau}\left(\frac{\mathrm dx}{\mathrm d\tau}\right) \\ &=& \frac{\mathrm d}{\mathrm d\tau} \left(\frac{\mathrm d t}{\mathrm d\tau}\frac{\mathrm dx}{\mathrm dt}\right) \\ &=& \frac{\mathrm d t}{\mathrm d\tau}\left(\frac{\mathrm d}{\mathrm d\tau}\frac{\mathrm dx}{\mathrm dt}\right) + \left(\frac{\mathrm d}{\mathrm d\tau} \frac{\mathrm d t}{\mathrm d\tau}\right)\frac{\mathrm dx}{\mathrm dt} \\ &=& \frac{\mathrm d t}{\mathrm d\tau}\left(\frac{\mathrm d t}{\mathrm d\tau}\frac{\mathrm d}{\mathrm dt}\frac{\mathrm dx}{\mathrm dt}\right) + \left(\frac{\mathrm d}{\mathrm d\tau} \frac{\mathrm d t}{\mathrm d\tau}\right)\frac{\mathrm dx}{\mathrm dt} \\ &=& \left(\frac{\mathrm d t}{\mathrm d\tau}\right)^2\frac{\mathrm d^2x}{\mathrm d t^2}+ \frac{\mathrm d^2 t}{\mathrm d\tau^2} \frac{\mathrm dx}{\mathrm d t}\;, \end{eqnarray}$$
which differs by the second term if $\frac{\mathrm d t}{\mathrm d\tau}$ is not constant.
I've just checked and $\frac{\mathrm d t}{\mathrm d\tau}$ is constant, but I still can't see how it works. Any chance of running through it step by step? Thank you. –  Peter4075 Sep 13 '11 at 17:22
@Peter4075: I thought that was step by step; could you say more specifically which part you're having trouble with? If $\frac{\mathrm d t}{\mathrm d\tau}$ is constant, then $\frac{\mathrm d^2 t}{\mathrm d\tau^2}=0$, and the second term vanishes, leaving the first term as expected. –  joriki Sep 13 '11 at 17:25
joriki - don't worry about making it too simple! I can see now about the second term disappearing if $\frac{dt}{d\tau}$ is constant. I can see how you get from the first line to the second line, but I'm afraid I still can't see how you get from the second line (a product) to the third line (a sum). –  Peter4075 Sep 13 '11 at 18:28
@Peter4075: I've tried to make it a bit more explicit -- hope that helps? The product turning into a sum is due to the product rule, $(fg)'=fg'+f'g$. –  joriki Sep 13 '11 at 19:02