2020 年研究生入学考试数学一第 10 题解析

题目

设 $\{\begin{array}{c}x=\sqrt{t^2+1}\\ y=\ln (t+\sqrt{t^2+1})\end{array}$, 则 $\frac{d^{2}y}{d{x}^2}{|}_{t=1}=$

解析

因为：

$$\begin{array}{r}\frac{dy}{dx}\\ & =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{array}$$

又：

$$\begin{array}{r}\frac{dy}{dt}\\ & =\frac{(t+\sqrt{t^2+1}{)}^{\prime }}{t+\sqrt{t^2+1}}\\ & =\frac{1}{t+\sqrt{t^2+1}}(1+\frac{t}{\sqrt{t^2+1}})\end{array}$$

$$\begin{array}{r}\frac{dx}{dt}\\ & =(\sqrt{t^2+1}{)}^{\prime }=[(t^2+1{)}^{\frac{1}{2}}{]}^{\prime }\\ & =\frac{t}{\sqrt{t^2+1}}\end{array}$$

于是：

$$\begin{array}{r}\frac{dy}{dx}\\ & =\frac{1}{t+\sqrt{t^2+1}}\times \frac{\sqrt{t^2+1}+t}{\sqrt{t^2+1}}\times \frac{\sqrt{t^2+1}}{t}\\ & =\frac{1}{t}\end{array}$$

又：

$$\begin{array}{r}\frac{d^{2}y}{d{x}^2}\\ & =\frac{d(\frac{dy}{dx})}{dx}\\ & =\frac{\frac{d(\frac{dy}{dx})}{dt}}{\frac{dx}{dt}}\end{array}$$

且：

$$\begin{array}{r}\frac{d(\frac{dy}{dx})}{dt}\\ & =(\frac{1}{t}{)}^{\prime }\\ & =\frac{-1}{t^2}\end{array}$$

于是：

$$\begin{array}{r}\frac{d^{2}y}{d{x}^2}\\ & =\frac{-1}{t^2}\times \frac{\sqrt{t^2+1}}{t}\\ & =\frac{-\sqrt{t^2+1}}{t^3}\end{array}$$

又：

$$t=1$$

于是：

$$\begin{array}{r}\frac{d^{2}y}{d{x}^2}\\ & =\frac{-\sqrt{2}}{1}\\ & =-\sqrt{2}\end{array}$$

EOF