一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
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 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
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:
$$
\left\{\begin{array}{l}f_{1}^{\prime}(1,0)+f_{2}^{\prime}(1,0)=3 \\ 2 f_{1}^{\prime}(1,0)+f_{2}^{\prime}(1,0)=5\end{array} \Rightarrow\right.
$$
$$
\left\{\begin{array}{l}f_{1}^{\prime}(1,0)=2 \\ f_{2}^{\prime}(1,0)=1\end{array}\right. \Rightarrow
$$
$$
\mathrm{d} f(1,0)=f_{1}^{\prime}(1,0) \mathrm{~d} x + f_{2}^{\prime}(1,0) \mathrm{~d} y =2 \mathrm{~d} x+\mathrm{~d} y
$$
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
以独特的视角解析线性代数,让繁复的知识变得直观明了。
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通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。