2021年考研数二第05题解析：泰勒公式、三角函数求导

一、题目

»A« $a = 1, b = – \frac{1}{2}$

»B« $a = 1, b = \frac{1}{2}$

»C« $a = 0, b = – \frac{1}{2}$

»D« $a = 0, b = \frac{1}{2}$

难度评级：

二、解析

首先，由泰勒公式的定义可知：

$$
\begin{aligned}
f\left(x\right) & = f\left(0\right) + f^{\prime}\left(0\right) x + \frac{f^{\prime\prime}\left(0\right)}{2!} x^{2} + o\left(x^{2}\right) \\ \\
& = f\left(0\right) + f^{\prime}\left(0\right) \cdot x + \frac{1}{2} \cdot f^{\prime\prime}\left(0\right) \cdot x^{2} + o\left(x^{2}\right)
\end{aligned}
$$

于是，当 $f\left(x\right) = \sec x$ 时，有：

$$
\begin{aligned}
f\left(0\right) & = \sec 0 = 1 \\ \\
f^{\prime}\left(0\right) & = \left.\left(\sec x \cdot \tan x\right)\right|_{x = 0} = 0 \\ \\
f^{\prime\prime}\left(0\right) & = \left.\left(\sec x \cdot \tan^{2} x + \sec^{3} x\right)\right|_{x = 0} = 1
\end{aligned}
$$

因此：

$$
f\left(x\right) = \sec x = 1 + \frac{1}{2} x^{2} + o\left(x^{2}\right)
$$

综上可知：

$$
\textcolor{lightgreen}{
a = 0, \ b = \dfrac{1}{2}
}
$$

综上可知，本 题 应 选 D

关于上面三角函数求导的过程，如果想要计算得比加快，就要首先记住下面两个公式：

$$
\begin{aligned}
\left( \sec x \right) ^{\prime} & = \textcolor{gray}{ \left( \frac{1}{\cos x} \right) ^{\prime} = \frac{\sin x}{\cos ^{2} x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} } = \tan x \cdot \sec x \\ \\
\left( \tan x \right) ^{\prime} & = \textcolor{gray}{ \left( \frac{\sin x}{\cos x} \right) ^{\prime} = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x} = \frac{1}{\cos^{2} x}} = \sec^{2} x
\end{aligned}
$$

于是：

$$
\begin{aligned}
\left( \sec x \cdot \tan x \right) ^{\prime} & = \left( \sec x \right) ^{\prime} \tan x + \sec x \left( \tan x \right) ^{\prime} \\ \\
& = \sec x \cdot \tan^{2} x + \sec^{3} x
\end{aligned}
$$

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