📝 题目
例 1 设 $f\left( {x,y}\right) = {x}^{2}{\mathrm{e}}^{y} + \left( {x - 1}\right) \arctan \frac{y}{x}$ ,求它在(1,0)点的偏导数.
💡 答案与解析
解法 1 因 $f\left( {x,0}\right) = {x}^{2}$ ,所以 ${f}_{x}^{\prime }\left( {1,0}\right) = 2$ . 同样因 $f\left( {1,y}\right) = {\mathrm{e}}^{y}$ , 所以 ${f}_{y}^{\prime }\left( {1,0}\right) = 1$ .
解法 2 因 ${f}_{x}^{\prime }\left( {x,y}\right) = {2x}{\mathrm{e}}^{y} + \arctan \frac{y}{x} + \frac{y\left( {1 - x}\right) }{{x}^{2} + {y}^{2}}$ ,所以 ${f}_{x}^{\prime }\left( {1,0}\right) = 2$ . 同样因 ${f}_{y}^{\prime }\left( {x,y}\right) = {x}^{2}{\mathrm{e}}^{y} + \frac{x\left( {x - 1}\right) }{{x}^{2} + {y}^{2}}$ ,得 ${f}_{y}^{\prime }\left( {1,0}\right) = 1$ .
可见求具体点的偏导数值时, 第一种解法较好.