早稲田大学 創造理工学研究科 経営システム工学専攻 2023年8月実施 微积分

标签：

• Waseda-University
• Implicit-Differentiation
• Differentiation

Author

思齐塾, 祭音Myyura

Description

$y$ が $x$ の関数 (function) であり, $e^{-x^2} + xy + \log y = 0$ を満たすとき, $y$ を $x$ で微分 (derivative) せよ. ただし, $y$ の自然対数 (natural logarithm) を $\log y$ として表現する.

题目描述

设 $y$ 是 $x$ 的函数，并满足

$$e^{-x^2}+xy+\log y=0,$$

其中 $\log y$ 表示 $y$ 的自然对数，因而 $y>0$。求 $\dfrac{dy}{dx}$。

Kai

Let $F(x, y) = e^{-x^2} + xy + \log y = 0$ . We want to find $\frac{dy}{dx}$ . Using implicit differentiation, we have:

$$\frac{d}{dx} (e^{-x^2} + xy + \log y) = 0$$

$$\frac{d}{dx} e^{-x^2} + \frac{d}{dx} (xy) + \frac{d}{dx} (\log y) = 0$$

Using the chain rule, we have $\frac{d}{dx} e^{-x^2} = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}$ . Using the product rule, we have $\frac{d}{dx} (xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y$ . Using the chain rule, we have $\frac{d}{dx} (\log y) = \frac{1}{y} \frac{dy}{dx}$ .

Substituting these into the equation, we get:

$$-2xe^{-x^2} + x \frac{dy}{dx} + y + \frac{1}{y} \frac{dy}{dx} = 0$$

Now, we solve for $\frac{dy}{dx}$ :

$$(x + \frac{1}{y}) \frac{dy}{dx} = 2xe^{-x^2} - y$$

$$\frac{dy}{dx} = \frac{2xe^{-x^2} - y}{x + \frac{1}{y}} = \frac{y(2xe^{-x^2} - y)}{xy + 1}$$

Therefore, at points where $xy+1\ne0$ ,

$$\boxed{\frac{dy}{dx}=\frac{2xye^{-x^2}-y^2}{xy+1}}.$$

The condition is necessary: since $\log y$ requires $y>0$ , if a point on the curve satisfies $xy+1=0$ , then

$$2xe^{-x^2}-y<0,$$

so the differentiated equation cannot have a finite value of $dy/dx$ there. Such a point has a vertical tangent rather than a finite derivative as a graph $y(x)$ .