设 $y=f(\ln x) e^{f(x)}$ ,其中 $f$ 二阶可导,求 $\mathrm{d} y$ 和 $y^{\prime \prime}(x)$.
【答案】 【参考解答】: 直接求导法则与复合函数求导法则,得
$$
y^{\prime}=f^{\prime}(\ln x) \cdot \frac{1}{x} \cdot e^{f(x)}+f(\ln x) e^{f(x)} f^{\prime}(x)
$$
所以得 $\mathrm{d} y=y^{\prime} \mathrm{d} x=e^{f(x)}\left[\frac{f^{\prime}(\ln x)}{x}+f(\ln x) f^{\prime}(x)\right] \mathrm{d} x$.
$$
\begin{aligned}
y^{\prime} &=\left[f^{\prime}(\ln x) \cdot \frac{1}{x} \cdot e^{f(x)}+f(\ln x) e^{f(x)} f^{\prime}(x)\right]^{\prime} \\
&=\frac{e^{f(x)} f^{\prime \prime}(\ln x)}{x^2}+e^{f(x)} f^{\prime \prime}(x) f(\ln x) \\
&-\frac{e^{f(x)} f^{\prime}(\ln x)}{x^2}+e^{f(x)} f^{\prime}(x)^2 f(\ln x)+\frac{2 e^{f(x)} f^{\prime}(x) f^{\prime}(\ln x)}{x}
\end{aligned}
$$


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