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$$$y^{2} = 10 x$$$ 关于 $$$x$$$ 的隐式导数
您的输入
求$$$\frac{d}{dx} \left(y^{2} = 10 x\right)$$$。
解答
分别对等式两边求导（将 $$$y$$$ 视为 $$$x$$$ 的函数）：$$$\frac{d}{dx} \left(y^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(10 x\right)$$$。
对方程的左边求导。
函数$$$y^{2}{\left(x \right)}$$$是两个函数$$$f{\left(u \right)} = u^{2}$$$和$$$g{\left(x \right)} = y{\left(x \right)}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)}$$应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = 2$$$:
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$返回到原变量：
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) = 2 {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$因此，$$$\frac{d}{dx} \left(y^{2}{\left(x \right)}\right) = 2 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$$。
对等式右边求导。
对 $$$c = 10$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(10 x\right)\right)} = {\color{red}\left(10 \frac{d}{dx} \left(x\right)\right)}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$10 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 10 {\color{red}\left(1\right)}$$因此，$$$\frac{d}{dx} \left(10 x\right) = 10$$$。
因此，我们得到如下关于导数的线性方程：$$$2 y \frac{dy}{dx} = 10$$$。
解得：$$$\frac{dy}{dx} = \frac{5}{y}$$$。
答案
$$$\frac{dy}{dx} = \frac{5}{y}$$$A