Question

6单选(2分)已知u=√R2-x2-y2-2,则全微分du=(A.dx+dy+dzVR2-x2-y2-22B.dx++dy+

Answer 1

问：6单选(2分)已知u=√R2-x2-y2-2,则全微分du=(A.dx+dy+dzVR2-x2-y2-22B.dx++dy+

答：你好，已知u=√R2-x2-y2-2,则全微分du=(A.dx+dy+dzVR2-x2-y2-22B.dx++dy。由题意：u = √(R^2 - x^2 - y^2 - 2)对于u的全微分du，可以利用偏导数进行求解：du = (∂u/∂x)dx + (∂u/∂y)dy + (∂u/∂z)dz其中，各偏导数计算如下：∂u/∂x = -x/(√(R^2 - x^2 - y^2 - 2))∂u/∂y = -y/(√(R^2 - x^2 - y^2 - 2))∂u/∂z = -1/(√(R^2 - x^2 - y^2 - 2))带入原式，得到：du = (-x/(√(R^2 - x^2 - y^2 - 2)))dx + (-y/(√(R^2 - x^2 - y^2 - 2)))dy - (dz/(√(R^2 - x^2 - y^2 - 2)))两边同时乘以分母(√(R^2 - x^2 - y^2 - 2))，得到：du = (-x/(R^2 - x^2 - y^2 - 2))dx + (-y/(R^2 - x^2 - y^2 - 2))dy - (dz/(R^2 - x^2 - y^2 - 2))故答案为：du = (-x/(R^2 - x^2 - y^2 - 2))dx + (-y/(R^2 - x^2 - y^2 - 2))dy - (dz/(R^2 - x^2 - y^2 - 2))