Question

设z等于sinxy➕Ln(x➕y)求偏导数

Answer 1

问：设z等于sinxy➕Ln(x➕y)求偏导数

答：我们需要分别对 $z=\sin(xy)+\ln(x+y)$ 对 $x$ 和 $y$ 求偏导数。首先对 $x$ 求偏导数，可以使用链式法则和求导法则：$$\begin{aligned}\frac{\partial z}{\partial x} &= \frac{\partial}{\partial x} \left[\sin(xy) + \ln(x+y)\right] \\&= \frac{\partial}{\partial x} \sin(xy) + \frac{\partial}{\partial x} \ln(x+y) \\&= y\cos(xy) + \frac{1}{x+y}\end{aligned}$$然后对 $y$ 求偏导数，同样使用链式法则和求导法则：$$\begin{aligned}\frac{\partial z}{\partial y} &= \frac{\partial}{\partial y} \left[\sin(xy) + \ln(x+y)\right] \\&= \frac{\partial}{\partial y} \sin(

答：$$然后对 $y$ 求偏导数，同样使用链式法则和求导法则：$$\begin{aligned}\frac{\partial z}{\partial y} &= \frac{\partial}{\partial y} \left[\sin(xy) + \ln(x+y)\right] \\&= \frac{\partial}{\partial y} \sin(xy) + \frac{\partial}{\partial y} \ln(x+y) \\&= x\cos(xy) + \frac{1}{x+y}\end{aligned}$$因此，$\frac{\partial z}{\partial x} = y\cos(xy) + \frac{1}{x+y}$，$\frac{\partial z}{\partial y} = x\cos(xy) + \frac{1}{x+y}$。