3.已知 e^z-sinxyz=xy 求(z)/(x) , (z)/(y)
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对于已知方程 $e^z - \sin(xy)z = xy$,要求 $\frac{z}{x}$ 和 $\frac{z}{y}$,可以利用偏导数的概念来求解。首先对方程两边同时求偏导数,得到:$\frac{\partial}{\partial x}(e^z - \sin(xy)z) = y$$\frac{\partial}{\partial y}(e^z - \sin(xy)z) = x$因此:$\frac{\partial z}{\partial x}e^z - z\cos(xy)\frac{\partial (xy)}{\partial x} = y$$\frac{\partial z}{\partial y}e^z - z\cos(xy)\frac{\partial (xy)}{\partial y} = x$化简原式,得到:$\frac{z}{x} = \frac{ye^z}{e^z - zy\cos(xy)}$$\frac{z}{y} = \frac{xe^z}{e^z - zx\cos(xy)}$因此,$\frac{z}{x} = \fr
咨询记录 · 回答于2023-06-08
3.已知 e^z-sinxyz=xy 求(z)/(x) , (z)/(y)
对于已知方程 $e^z - \sin(xy)z = xy$,要求 $\frac{z}{x}$ 和 $\frac{z}{y}$,可以利用偏导数的概念来求解。首先对方程两边同时求偏导数,得到:$\frac{\partial}{\partial x}(e^z - \sin(xy)z) = y$$\frac{\partial}{\partial y}(e^z - \sin(xy)z) = x$因此:$\frac{\partial z}{\partial x}e^z - z\cos(xy)\frac{\partial (xy)}{\partial x} = y$$\frac{\partial z}{\partial y}e^z - z\cos(xy)\frac{\partial (xy)}{\partial y} = x$化简原式,得到:$\frac{z}{x} = \frac{ye^z}{e^z - zy\cos(xy)}$$\frac{z}{y} = \frac{xe^z}{e^z - zx\cos(xy)}$因此,$\frac{z}{x} = \fr
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