做了这道题你会对全微分有更深入的理解

一、题目

已知函数 $f(x, y)$ 可微，且 $f[x+1, \ln (1+x)]$ $=$ $(1+x)^{3}+x \ln (1+x)(x+1)^{\ln (x+1)}$, $f\left(x^{2}, x-1\right)$ $=$ $x^{4} \mathrm{e}^{x-1}+(x-1)\left(x^{2}-1\right) x^{2(x-1)}$.

则：

$$
\mathrm{d} f(1,0) = ?
$$

难度评级：

二、解析

令：

$$
g(x)=x \ln (1+x)(x+1)^{\ln (x+1)}
$$

$$
k(x)=(x-1)\left(x^{2}-1\right) x^{2(x-1)}
$$

则：

$$
f[x+1, \ln (1+x)]=(1+x)^{3}+g(x) \Rightarrow
$$

$$
df[x+1, \ln (1+x)] = \mathrm{~d} \left[(1+x)^{3}+g(x)\right] \Rightarrow
$$

$$
\left.f_{1}^{\prime}[x+1, \ln (1+x)] \cdot 1+f_{2}^{\prime}[x+1, \ln (1+x)\right] \cdot \frac{1}{1+x} =
$$

$$
3(1+x)^{2}+g^{\prime}(x) \Rightarrow
$$

$$
\left\{\begin{array}{l}x=1 \\ y=0\end{array} \Rightarrow \left\{\begin{array}{l}x+1=1 \\ \ln(1+x)=0 \end{array} \Rightarrow x =0\right.\right. \Rightarrow
$$

$$
f_{1}^{\prime}(1,0)+f_{2}^{\prime}(1,0) \cdot \frac{1}{1+0} =
$$

$$
3(1+0)^{2}+g^{\prime}(0)=3+g^{\prime}(0).
$$

又：

$$
g^{\prime}(0) = \lim \limits_{x \rightarrow 0} \frac{g(x)-g(0)}{x-0}=\lim \limits_{x \rightarrow 0} \ln (1+x)(x+1)^{\ln (x+1)}=
$$

$$
0 \cdot 1^{0}=0 \cdot 1=0.
$$

于是：

$$
f_{1}^{\prime}(1,0)+f_{2}^{\prime}(1,0) = 3.
$$

接着：

$$
f\left(x^{2}, x-1\right)=x^{4} e^{x-1}+k(x) \Rightarrow
$$

$$
\mathrm{d} f\left(x^{2}, x-1\right)=
$$

$$
\mathrm{~d} \left[x^{4} e^{x-1}+k(x)\right] \Rightarrow
$$

$$
f_{1}^{\prime}\left(x^{2}, x-1\right) \cdot 2 x+f_{2}^{\prime}\left(x^{2}, x-1\right) \cdot 1=
$$

$$
4 x^{3} e^{x-1}+x^{4} e^{x-1}+k^{\prime}(x).
$$

又：

$$
\left\{\begin{array}{l}x=1 \\ y=0\end{array} \Rightarrow\left\{\begin{array}{l}x^{2}=1 \\ x-1=0\end{array} \Rightarrow x=1\right.\right.
$$

于是：

$$
f_{1}^{\prime}(1,0) \cdot 2+f_{2}^{\prime}(1,0)=4+1+k^{\prime}(0) =5+k^{\prime}
$$

$$
k^{\prime}(1)=\lim \limits_{x \rightarrow 1} \frac{k(x)-k(1)}{x-1}=\lim \limits_{x \rightarrow 1} \frac{k(x)}{x-1}=
$$

$$
\lim \limits_{x \rightarrow 1}\left(x^{2}-1\right) x^{2(x-1)}=0 \cdot 1^{0}=0 \cdot 1=0
$$

综上：

$$
\left\{\begin{array}{l}f_{1}^{\prime}(1,0)+f_{2}^{\prime}(1,0)=3 \\ 2 f_{1}^{\prime}(1,0)+f_{2}^{\prime}(1,0)=5\end{array} \Rightarrow\right.
$$

$$
\left\{\begin{array}{l}f_{1}^{\prime}(1,0)=2 \\ f_{2}^{\prime}(1,0)=1\end{array}\right. \Rightarrow
$$

$$
\mathrm{d} f(1,0)=f_{1}^{\prime}(1,0) \mathrm{~d} x + f_{2}^{\prime}(1,0) \mathrm{~d} y =2 \mathrm{~d} x+\mathrm{~d} y
$$

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