问题及解答

设 $f(x)$ 二阶可导,  求下列函数的二阶导数.

(1) $y=f(\frac{1}{x})$

(2) $y=e^{f(x)}$

(1) $y'=f'(\frac{1}{x})\cdot\frac{-1}{x^2}=-\frac{1}{x^2}f'(\frac{1}{x})$,

\[
\begin{split}
y''&=-\Bigl[(\frac{1}{x^2})'\cdot f'(\frac{1}{x})+\frac{1}{x^2}\cdot f''(\frac{1}{x})\cdot\frac{-1}{x^2}\Bigr]\\
&=-\Bigl[(-2)x^{-3}\cdot f'(\frac{1}{x})-\frac{1}{x^4}\cdot f''(\frac{1}{x})\Bigr]\\
&=\frac{2}{x^3} f'(\frac{1}{x})+\frac{1}{x^4}f''(\frac{1}{x}).
\end{split}
\]

(2)

\[
y'=e^{f(x)}\cdot f'(x)=f'(x)e^{f(x)},
\]

\[
y''=f''(x)e^{f(x)}+f'(x)(e^{f(x)})'=f''(x)e^{f(x)}+(f'(x))^2\cdot e^{f(x)}.
\]