2021年考研数二第12题解析：参数方程求导

一、题目

难度评级：

二、解析

首先：

$$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{d} x} & = \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{\frac{\mathrm{d} x}{\mathrm{d} t}} \\ \\
\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} & = \frac{\mathrm{d} \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)}{\mathrm{d} x} = \frac{\mathrm{d} \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} = \frac{\mathrm{d} \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)}{\mathrm{d} t} \cdot \frac{1}{\frac{\mathrm{d} x}{\mathrm{d} t}}
\end{aligned}
$$

又因为：

$$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{d} t} & = 4 \mathrm{e}^{t} + 4 \left( t-1 \right) \mathrm{e}^{t} + 2t \\ \\
& = 4 t \mathrm{e}^{t} + 2t \\ \\ \\
\frac{\mathrm{d} x}{\mathrm{d} t} & = 2 \mathrm{e}^{t} + 1
\end{aligned}
$$

于是：

$$
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4t\mathrm{e}^{t}+2t}{2\mathrm{e}^{t}+1} = \frac{2t \left( 2 \mathrm{e}^{t} + 1 \right)}{2 \mathrm{e}^{t} + 1} = 2t
$$

进而：

$$
\frac{\mathrm{d} \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)}{\mathrm{d} t} = \left( 2t \right) ^{\prime} = 2
$$

综上，当 $t = 0$ 时，有：

$$
\left.\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right|_{t=0} = 2 \cdot \frac{1}{2+1} = \frac{2}{3}
$$

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