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<h1>Solución</h1>
<p>Sabemos que
$\frac{{dy}}{{dx}} = \frac{2}{{2t}} = \frac{1}{t}$</p>
<p>Usando la ecuación, obtenemos</p>

$$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{dx/dt}} = \frac{{\frac{d}{{dt}}\left( {\frac{1}{t}} \right)}}{{2t}} = \frac{{ - {t^{ - 2}}}}{{2t}} = \frac{{ - 1}}{{2{t^3}}}$$

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