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From: ViperEkura <3081035982@qq.com>
Date: Tue, 7 Apr 2026 18:46:13 +0800
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+# 隐函数求导
+
+
+## 对于一阶求导方程
+给定函数 $F(x,y) = 0$
+
+$\frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx}$ ，推导得到 $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0$
+
+## 对于二阶或以上求导（以二阶为例）
+
+给定函数方程 $A(x,y)\frac{dy}{dx} + B(x,y) = 0$
+
+$$
+\frac{dA}{dx} \cdot \frac{dy}{dx} + A(x,y)\frac{d^2y}{dx^2} + \frac{dB}{dx} = 0
+$$
+
+其中：
+$$\begin{cases} 
+\frac{dA}{dx} = \frac{\partial A}{\partial x} + \frac{\partial A}{\partial y} \cdot \frac{dy}{dx} \\ 
+\frac{dB}{dx} = \frac{\partial B}{\partial x} + \frac{\partial B}{\partial y} \cdot \frac{dy}{dx} 
+\end{cases}$$
+
+**同理可推广至 $n$ 阶**
+
+设 $F(x, y) = 0$ 确定隐函数 $y = y(x)$，对 $x$ 求 $n$ 阶导数：
+$$\sum_{k=0}^{n} \binom{n}{k} \frac{\partial^{n-k} F}{\partial x^{n-k}} \cdot \frac{d^k y}{dx^k} = 0$$
+
+即：
+$$\frac{\partial^n F}{\partial x^n} + \binom{n}{1} \frac{\partial^{n-1} F}{\partial x^{n-1} \partial y} \cdot \frac{dy}{dx} + \cdots + \frac{\partial F}{\partial y} \cdot \frac{d^n y}{dx^n} = 0$$
+
+可解出：
+$$\frac{d^n y}{dx^n} = -\frac{1}{\frac{\partial F}{\partial y}} \left( \frac{\partial^n F}{\partial x^n} + \sum_{k=1}^{n-1} \binom{n}{k} \frac{\partial^{n-k} F}{\partial x^{n-k} \partial y} \cdot \frac{d^k y}{dx^k} \right)$$
+
+---
+
+## 参数方程求导
+
+参数方程设定为 $\begin{cases} x = \phi(t) \\ y = \psi(t) \end{cases}$
+
+
+$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta t \to 0} \frac{\Delta y / \Delta t}{\Delta x / \Delta t} = \frac{dy/dt}{dx/dt} = \frac{\psi'(t)}{\phi'(t)}$
+
+$\Delta t \to 0$ 时 $\Delta x \to 0$。
+
+## 对于二阶或以上的范围进行求导
+
+$\frac{d^2y}{dx^2} = \frac{d(\frac{dy}{dx})}{dx} = \frac{d(\frac{dy}{dx})/dt}{dx/dt} = \frac{\psi''(t)\phi'(t) - \psi'(t)\phi''(t)}{(\phi'(t))^3}$
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