Answer
Equation (13) is obtained by using the chain rule and the equations:
$\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^2}}}$
and
$\frac{{dt}}{{dx}} = \frac{1}{{x'\left( t \right)}}$
Work Step by Step
From Theorem 2 we have the slope of the tangent line:
$\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}}$
Taking the derivative of $\frac{{dy}}{{dx}}$ with respect to $t$ gives
$\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dt}}\left( {\frac{{y'\left( t \right)}}{{x'\left( t \right)}}} \right) = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^2}}}$
Next, we evaluate $\frac{{{d^2}y}}{{d{x^2}}}$ by using the chain rule of derivatives:
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)\frac{{dt}}{{dx}}$
Using $\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^2}}}$ and $\frac{{dt}}{{dx}} = \frac{1}{{x'\left( t \right)}}$ we obtain the equation (13):
(13) $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}}$
