题目
设 $\left\{\begin{matrix} x = \sqrt{t^{2}+1} \\ y = \ln(t+\sqrt{t^{2}+1}) \end{matrix}\right.$, 则 $\frac{d^{2}y}{dx^{2}}|_{t=1}=$
解析
因为：
\begin{align}
\frac{dy}{dx} \\
& = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\end{align}
又：
\begin{align}
\frac{dy}{dt} \\
& =\frac{(t+\sqrt{t^{2}+1})’}{t+\sqrt{t^{2}+1}} \\
& = \frac{1}{t+\sqrt{t^{2}+1}}(1+\frac{t}{\sqrt{t^{2}+1}})
\end{align}
\begin{align}
\frac{dx}{dt} \\
& = (\sqrt{t^{2}+1})’ \ =[(t^{2}+1)^{\frac{1}{2}}]’ \\
& = \frac{t}{\sqrt{t^{2}+1}}
\end{align}
于是：
\begin{align}
\frac{dy}{dx} \\
& =\frac{1}{t+\sqrt{t^{2}+1}} \times \frac{\sqrt{t^{2}+1}+t}{\sqrt{t^{2}+1}} \times \frac{\sqrt{t^{2}+1}}{t} \\
& = \frac{1}{t}
\end{align}
又：
\begin{align}
\frac{d^{2}y}{dx^{2}} \\
& = \frac{d(\frac{dy}{dx})}{dx} \\
& = \frac{\frac{d(\frac{dy}{dx})}{dt}}{\frac{dx}{dt}}
\end{align}
且：
\begin{align}
\frac{d(\frac{dy}{dx})}{dt} \\
& = (\frac{1}{t})’ \\
& = \frac{-1}{t^{2}}
\end{align}
于是：
\begin{align}
\frac{d^{2}y}{dx^{2}} \\
& = \frac{-1}{t^{2}} \times \frac{\sqrt{t^{2}+1}}{t} \\
& = \frac{-\sqrt{t^{2}+1}}{t^{3}}
\end{align}
又：
$$t=1$$
于是：
\begin{align}
\frac{d^{2}y}{dx^{2}} \\
& = \frac{-\sqrt{2}}{1} \\
& = -\sqrt{2}\end{align}
EOF