链式法则

一阶导

令 $y=f(u)$, $u=g(x)$, 即 $y = f(g(x))$, 则

$$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$$

二阶导

令 $y=f(u)$, $u=g(x)$, 即 $y = f(g(x))$, 则

$$\begin{align} \dfrac{d^2 y}{dx^2} &= \dfrac{d}{dx}(\dfrac{dy}{dx}) \\ &= \dfrac{d}{dx}(\dfrac{dy}{du} \cdot \dfrac{du}{dx}) \\ &= \dfrac{1}{dx}(\dfrac{d^2 y}{du} \cdot \dfrac{du}{dx} + \dfrac{dy}{du} \cdot \dfrac{d^2 u}{dx}) \text{(微分乘法法则)} \\ &= \dfrac{d^2 y}{du} \cdot \dfrac{du}{dx^2} + \dfrac{dy}{du} \cdot \dfrac{d^2 u}{dx^2} \\ &= \dfrac{d^2 y}{du^2} \cdot \dfrac{du^2}{dx^2} + \dfrac{dy}{du} \cdot \dfrac{d^2 u}{dx^2} \text{(配凑du)} \\ &= \dfrac{d^2 y}{du^2} \cdot (\dfrac{du}{dx})^2 + \dfrac{dy}{du} \cdot \dfrac{d^2 u}{dx^2} \\ \end{align}$$

警告

易混点: $\dfrac{d^2 y}{dx}$ 为二阶导, $\dfrac{dy^2}{dx^2} = (\dfrac{dy}{dx})^2$ 为一阶导的平方; 计算 $\dfrac{df}{dx}$ 时须先按照微分法则转换 $df$, 而非 $\dfrac{df}{dx}$ 整体