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?
 A: 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.
