确定一点处是否可导？直接用一点处导数的定义试试看吧！

一、题目

下列函数中在 $x=0$ 处可导的是哪个或哪些？

(1) $f(x)=\cos x^{\frac{2}{3}}$;

(2) $f(x)=\sin x^{\frac{2}{3}}$;

(3) $f(x)=(1-\cos x)^{\frac{2}{3}}$;

(4) $f(x)=\left(\sin x^{2}\right)^{\frac{1}{3}}$.

难度评级：

二、解析

(1):

$$
f^{\prime}(0)=\lim \limits_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x}=\lim \limits_{x \rightarrow 0} \frac{\cos x^{\frac{2}{3}}-1}{x}=
$$

$$
\lim \limits_{x \rightarrow 0} \frac{-\frac{1}{2}\left(x^{\frac{2}{3}}\right)^{2}}{x}=\lim \limits_{x \rightarrow 0} \frac{-1}{2} x^{\frac{1}{3}}=0 \Rightarrow 可导
$$

(2):

$$
f^{\prime}(0)=\lim \limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim \limits_{x \rightarrow 0} \frac{\sin x^{\frac{2}{3}}-0}{x}=
$$

$$
\lim \limits_{x \rightarrow 0} \frac{x^{\frac{2}{3}}}{x}=\lim \limits_{x \rightarrow 0} x^{\frac{-1}{3}}=\lim \limits_{x \rightarrow 0} \frac{1}{x^{\frac{1}{3}}}=\infty \Rightarrow 不可导
$$

(3):

$$
f^{\prime}(0)=\lim \limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim \limits_{x \rightarrow 0} \frac{(1-\cos x)^{\frac{2}{3}}-0}{x}=
$$

$$
\lim \limits_{x \rightarrow 0} \frac{\frac{1}{2}\left(x^{2}\right)^{\frac{2}{3}}}{x}=\lim \limits_{x \rightarrow 0} \frac{1}{2} x^{\frac{1}{3}}=0 \Rightarrow 可导
$$

(4):

$$
f^{\prime}(0)=\lim \limits_{x \rightarrow 0} \frac{\left(\sin x^{2}\right)^{\frac{1}{3}}-0}{x}=\lim \limits_{x \rightarrow 0} \frac{x^{\frac{2}{3}}}{x}=
$$

$$
\lim \limits_{x \rightarrow 0} x^{\frac{-1}{3}}=\lim \limits_{x \rightarrow 0} \frac{1}{x^{\frac{1}{3}}}=\infty \Rightarrow 不可导
$$

综上，可导的是 (1) 和 (3).

---