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按来源筛选并翻页浏览已保存的题目记录。

#1选择题来源:301-2026
设函数 $z = z(x,y)$ 由方程 $x - az = e^{y+az}$ (a是非0常数)确定,则 ( )
  • A. $\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = \dfrac{1}{a}$
  • B. $\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} = \dfrac{1}{a}$
  • C. $\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = -\dfrac{1}{a}$
  • D. $\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} = -\dfrac{1}{a}$
#1选择题来源:302-2025
设函数 $z=z(x,y)$ 由 $z+\ln z-\int_{y}^{x} e^{-t^{2}}\,dt=0$ 确定,则 $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=(\ )$
  • A. $\dfrac{z}{z+1}\left(e^{-x^{2}}-e^{-y^{2}}\right)$
  • B. $\dfrac{z}{z+1}\left(e^{-x^{2}}+e^{-y^{2}}\right)$
  • C. $-\dfrac{z}{z+1}\left(e^{-x^{2}}-e^{-y^{2}}\right)$
  • D. $-\dfrac{z}{z+1}\left(e^{-x^{2}}+e^{-y^{2}}\right)$
#2选择题来源:303-2026
设函数 $z = z(x,y)$ 由方程 $x - az = e^{y+az}$($a$ 是非零常数)确定,则 ( )
  • A. $\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = \dfrac{1}{a}$
  • B. $\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} = \dfrac{1}{a}$
  • C. $\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = -\dfrac{1}{a}$
  • D. $\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} = -\dfrac{1}{a}$
#3选择题来源:302-2026
设函数 $z = z(x,y)$ 由方程 $x - az = e^{y+ax}$($a$ 是非零常数)确定,则 ( )
  • A. $\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = \dfrac{1}{a}$
  • B. $\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} = \dfrac{1}{a}$
  • C. $\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = -\dfrac{1}{a}$
  • D. $\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} = -\dfrac{1}{a}$
#13填空题来源:302-2021
设函数 $z=z(x,y)$ 由方程 $(x+1)z+y\ln z-\arctan 2xy=1$ 确定,则 $\left.\frac{\partial z}{\partial x}\right|_{(0,2)}=\underline{\qquad}$.
#13填空题来源:302-2023
设函数 $z=z(x,y)$ 由 $e^{z}+xz=2x-y$ 确定,则 $\left.\frac{\partial^{2}z}{\partial x^{2}}\right|_{(1,1)}=\underline{\qquad}$.
#18解答题来源:303-2024
(本题满分12分) 设函数$z = z(x,y)$由方程$z + \mathrm{e}^x - y\ln(1 + z^2) = 0$确定,求$\left.\left(\dfrac{\partial^2 z}{\partial x^2} + \dfrac{\partial^2 z}{\partial y^2}\right)\right|_{(0,0)}$。