2022考研数二第12题解析：隐函数求导

一、题目

二、解析

$\textcolor{lightgreen}{\blacktriangleright}$ 首先，将 $x = 1$ 代入方程 $x^{2} + x y + y^{3} = 3$, 得：

$$
\begin{aligned}
& \ 1 + y + y^{3} = 3 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ y = 1 }
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 接着，在方程 $x^{2} + x y + y^{3} = 3$ 等号两端同时对 $x$ 求导，得：

$$
\textcolor{lightgreen}{
2 x + y + x y^{\prime} + 3 y^{2} \cdot y^{\prime} = 0 } \tag{1}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 将 $x = 1, y = 1$ 代入上面的 $(1)$, 得：

$$
\begin{aligned}
& \ 2 + 1 + y ^{\prime} + 3 y ^{\prime} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 4 y ^{\prime} = -3 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ y^{\prime} = -\frac{3}{4} }
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 对 $(1)$ 式两边求导，得：

$$
\textcolor{lightgreen}{
2 + y^{\prime} + y^{\prime} + x y^{\prime\prime} + 6 y \cdot y^{\prime} \cdot y^{\prime} + 3 y^{2} \cdot y^{\prime\prime} = 0
} \tag{2}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 将 $y = 1, x = 1, y^{\prime} = -\frac{3}{4}$ 代入上面的 $(2)$ 式，得：

$$
\begin{aligned}
& \ 2 – \frac{3}{4} – \frac{3}{4} + y ^{\prime \prime} \left( 1 \right) + 6 \cdot \frac{3}{4} \cdot \frac{3}{4} + 3 y ^{\prime \prime} \left( 1 \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{1}{2} + \frac{27}{8} + 4 y ^{\prime \prime} \left( 1 \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 4 y ^{\prime \prime} \left( 1 \right) = \frac{-31}{8} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ y^{\prime\prime}\left(1\right) = -\frac{31}{32} }
\end{aligned}
$$

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