问题及解答

设 $y=f(x)$ 具有反函数, 其反函数为 $g(x)$. 并且 $g$ 二阶可导, 且满足 $(g'(x))^2=g''(x)$, 证明 $f''(x)+f(x)=0$.

$y=f(x)$ 的反函数写为 $x=g(y)$. 则

\[g'(y)=x'_y=\frac{1}{y'_x}=\frac{1}{f'(x)}\]

\[
g''(y)=(\frac{1}{f'(x)})'_y=\frac{-1}{(f'(x))^2}\cdot f''(x)\cdot x'_y=\frac{-1}{(f'(x))^2}\cdot f''(x)\cdot\frac{1}{f'(x)}
\]

由条件, $(g'(y))^2=g''(y)$, 得

\[
(\frac{1}{f'(x)})^2=\frac{-f''(x)}{(f'(x))^3}.
\]

此即推出 $f''(x)+f(x)=0$. 证毕.