[题目]-|||-求由参数方程 及二阶导数-|||-dfrac {{d)^2y}(d{x)^2}

步骤 1：求一阶导数 $\dfrac {dy}{dx}$
首先，我们分别对 $x$ 和 $y$ 关于 $t$ 求导。
$$
\dfrac {dx}{dt} = \dfrac {d}{dt} \ln \sqrt {1+{t}^{2}} = \dfrac {1}{\sqrt {1+{t}^{2}}} \cdot \dfrac {1}{2} \cdot \dfrac {2t}{\sqrt {1+{t}^{2}}} = \dfrac {t}{1+{t}^{2}}
$$
$$
\dfrac {dy}{dt} = \dfrac {d}{dt} \arctan t = \dfrac {1}{1+{t}^{2}}
$$
然后，根据链式法则，我们有
$$
\dfrac {dy}{dx} = \dfrac {\dfrac {dy}{dt}}{\dfrac {dx}{dt}} = \dfrac {\dfrac {1}{1+{t}^{2}}}{\dfrac {t}{1+{t}^{2}}} = \dfrac {1}{t}
$$

步骤 2：求二阶导数 $\dfrac {{d}^{2}y}{d{x}^{2}}$
首先，我们对 $\dfrac {dy}{dx}$ 关于 $t$ 求导。
$$
\dfrac {d}{dt} \left( \dfrac {dy}{dx} \right) = \dfrac {d}{dt} \left( \dfrac {1}{t} \right) = -\dfrac {1}{{t}^{2}}
$$
然后，根据链式法则，我们有
$$
\dfrac {{d}^{2}y}{d{x}^{2}} = \dfrac {d}{dx} \left( \dfrac {dy}{dx} \right) = \dfrac {d}{dt} \left( \dfrac {dy}{dx} \right) \cdot \dfrac {dt}{dx} = -\dfrac {1}{{t}^{2}} \cdot \dfrac {1}{\dfrac {dx}{dt}} = -\dfrac {1}{{t}^{2}} \cdot \dfrac {1+{t}^{2}}{t} = -\dfrac {1+{t}^{2}}{{t}^{3}}
$$