问题及解答

求下列分段函数的导数

(1)

\[
f(x)=\begin{cases}
\frac{\sqrt{x^2+1}-1}{x},& x\neq 0,\\
0,& x=0.
\end{cases}
\]

当 $x\neq 0$ 时,

\[
\begin{split}
f'(x)&=\bigl(\frac{\sqrt{1+x^2}-1}{x}\bigr)'=\frac{\frac{2x}{2\sqrt{1+x^2}}\cdot x-(\sqrt{1+x^2}-1)\cdot 1}{x^2}\\
&=\frac{x^2-\sqrt{1+x^2}(\sqrt{1+x^2}-1)}{x^2\cdot\sqrt{1+x^2}}\\
&=\frac{x^2-(1+x^2)+\sqrt{1+x^2}}{x^2\cdot\sqrt{1+x^2}}\\
&=\frac{\sqrt{1+x^2}-1}{x^2\cdot\sqrt{1+x^2}}\\
&=\frac{1}{\sqrt{1+x^2}(\sqrt{1+x^2}+1)}.
\end{split}
\]

当 $x=0$ 时,

\[
\begin{split}
f'(0)&=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}\\
&=\lim_{x\rightarrow 0}\frac{\frac{\sqrt{1+x^2}-1}{x}-0}{x}\\
&=\lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-1}{x^2}\\
&=\lim_{x\rightarrow 0}\frac{1}{\sqrt{1+x^2}+1}\\
&=\frac{1}{2}.
\end{split}
\]