早稲田大学 創造理工学研究科 経営システム工学専攻 2013年8月実施 微积分

标签：

• Waseda-University
• Differentiation

Author

思齐塾, 祭音Myyura

Description

$x>0$ に対し、次の関数 (function) の微分 (derivative) を求めよ.

$$f(x) = \left(1 + \frac{1}{x}\right)^x$$

题目描述

设 $x>0$。求函数

$$f(x)=\left(1+\frac{1}{x}\right)^x$$

的导数。

Kai

Let $y = f(x) = (1 + \frac{1}{x})^x$ . Then, take the natural logarithm of both sides:

$$\ln y = \ln \left( \left(1 + \frac{1}{x}\right)^x \right) = x \ln \left(1 + \frac{1}{x}\right)$$

Now, differentiate both sides with respect to $x$ :

$$\frac{1}{y} \frac{dy}{dx} = \ln \left(1 + \frac{1}{x}\right) + x \cdot \frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right)$$

$$\frac{1}{y} \frac{dy}{dx} = \ln \left(1 + \frac{1}{x}\right) + x \cdot \frac{x}{x+1} \cdot \left(-\frac{1}{x^2}\right)$$

$$\frac{1}{y} \frac{dy}{dx} = \ln \left(1 + \frac{1}{x}\right) - \frac{x}{x+1} \cdot \frac{1}{x}$$

$$\frac{1}{y} \frac{dy}{dx} = \ln \left(1 + \frac{1}{x}\right) - \frac{1}{x+1}$$

Now, multiply both sides by $y$ to find $\frac{dy}{dx}$ :

$$\frac{dy}{dx} = y \left[\ln \left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right]$$

Substitute back the expression for $y$ :

$$\frac{dy}{dx} = \left(1 + \frac{1}{x}\right)^x \left[\ln \left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right]$$

So, the derivative is:

$$f'(x) = \left(1 + \frac{1}{x}\right)^x \left[\ln \left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right]$$