题目
已知 z = f(x, xy, x^2 + y^2) 具有连续偏导数,则下面正确的是()。A. (partial z)/(partial x) = f_1';B. (partial z)/(partial x) = f_1' + yf_2' + 2xf_3';C. (partial z)/(partial x) = f_1' + f_2' + f_3';D. (partial z)/(partial x) = f_1' + xf_2' + 2yf_3'。
已知 $z = f(x, xy, x^2 + y^2)$ 具有连续偏导数,则下面正确的是()。
A. $\frac{\partial z}{\partial x} = f_1'$;
B. $\frac{\partial z}{\partial x} = f_1' + yf_2' + 2xf_3'$;
C. $\frac{\partial z}{\partial x} = f_1' + f_2' + f_3'$;
D. $\frac{\partial z}{\partial x} = f_1' + xf_2' + 2yf_3'$。
题目解答
答案
B. $\frac{\partial z}{\partial x} = f_1' + yf_2' + 2xf_3'$;
解析
本题考查多元复合函数求偏导数的链式法则。解题思路是先设出中间变量,将原函数表示为关于中间变量的复合函数,然后根据链式法则列出求偏导数的公式,再分别计算中间变量对自变量的偏导数,最后代入公式得到结果。
设 $u = x$,$v = xy$,$w = x^2 + y^2$,则 $z = f(u, v, w)$。
根据多元复合函数求偏导数的链式法则,若 $z = f(u, v, w)$,$u = u(x,y)$,$v = v(x,y)$,$w = w(x,y)$,则$\frac{\partial z}{\partial x} = f_1' \cdot \frac{\partial u}{\partial x} + f_2' \cdot \frac{\partial v}{\partial x} + f_3' \cdot \frac{\partial w}{\partial x}$。
接下来分别计算$\frac{\partial u}{\partial x}$,$\frac{\partial v}{\partial x}$,$\frac{\partial w}{\partial x}$:
- 对于$u = x$,根据求导公式$(x^n)^\prime=nx^{n - 1}$,这里$n = 1$,可得$\frac{\partial u}{\partial x} = 1$。
- 对于$v = xy$,把$y$看作常数,对$x$求偏导数,根据求导公式$(ax)^\prime=a$($a$为常数),可得$\frac{\partial v}{\partial x} = y$。
- 对于$w = x^2 + y^2$,把$y$看作常数,对$x$求偏导数,根据求导公式$(x^n)^\prime=nx^{n - 1}$,可得$\frac{\partial w}{\partial x} = 2x$。
将$\frac{\partial u}{\partial x} = 1$,$\frac{\partial v}{\partial x} = y$,$\frac{\partial w}{\partial x} = 2x$代入$\frac{\partial z}{\partial x} = f_1' \cdot \frac{\partial u}{\partial x} + f_2' \cdot \frac{\partial v}{\partial x} + f_3' \cdot \frac{\partial w}{\partial x}$,可得:
$\frac{\partial z}{\partial x} = f_1' \cdot 1 + f_2' \cdot y + f_3' \cdot 2x = f_1' + yf_2' + 2xf_3'$