(14)设 y=y(x) 由 ^ysin t-y+1=0 和 x= { ,tneq 0 t,t=0|=t=0 __

步骤 1：求解 $x'(t)$ 和 $x''(t)$
$x(t)=\left \{ \begin{matrix} \dfrac {{e}^{t}-1}{t},\quad t\neq 0,\\ 1,\quad t=0\end{matrix} \right.$
当 $t\neq 0$ 时，$x'(t)=\dfrac {t{e}^{t}-({e}^{t}-1)}{{t}^{2}}=\dfrac {(t-1){e}^{t}+1}{{t}^{2}}$
$x''(t)=\dfrac {t^{2}[(t-1){e}^{t}+{e}^{t}]-2t[(t-1){e}^{t}+1]}{t^{4}}=\dfrac {t^{2}{e}^{t}-2t{e}^{t}-2}{t^{3}}$
当 $t=0$ 时，$x'(0)=\lim_{t\to 0}\dfrac {x(t)-x(0)}{t-0}=\lim_{t\to 0}\dfrac {e^{t}-1}{t^{2}}=\lim_{t\to 0}\dfrac {e^{t}}{2t}=\dfrac {1}{2}$
$x''(0)=\lim_{t\to 0}\dfrac {x'(t)-x'(0)}{t-0}=\lim_{t\to 0}\dfrac {(t-1)e^{t}+1}{t^{2}}-\dfrac {1}{2}=\lim_{t\to 0}\dfrac {e^{t}-1}{2t}=\dfrac {1}{3}$
步骤 2：求解 $y'(t)$ 和 $y''(t)$
由 ${e}^{y}\sin t-y+1=0$，对 $t$ 求导得 ${e}^{y}y'\sin t+{e}^{y}\cos t-y'=0$
当 $t=0$ 时，$y(0)=1$，$y'(0)=e$
对 $t$ 求导得 ${e}^{y}y'\sin t+{e}^{y}y'\cos t+{e}^{y}\cos t-{e}^{y}\sin t-y''=0$
当 $t=0$ 时，$y''(0)=2e^{2}$
步骤 3：求解 $\dfrac {{d}^{2}y}{d{x}^{2}}{|}_{t=0}$
$\dfrac {{d}^{2}y}{d{x}^{2}}{|}_{t=0}=\dfrac {y'(0)x'(0)-x''(0)y'(0)}{x'(0)^{3}}=\dfrac {e\times \dfrac {1}{2}-\dfrac {1}{3}e}{(\dfrac {1}{2})^{3}}=8e(e-\dfrac {1}{3})$