Question

高等数学偏导数？

如图，这个偏导数要怎么做？求下过程，谢谢啦

Answer 1

Answer 2

Answer 3

xe^x - ye^y = ze^z
两边对 x 求偏导，得 e^x + x^e^x = (e^z+ze^z)∂z/∂x
则 ∂z/∂x = (e^x+xe^x)/(e^z+ze^z),
同理，∂z/∂y = -(e^y+ye^y)/(e^z+ze^z)
dz = (∂z/∂x)dx + (∂z/∂y)dy
= (e^x+xe^x)dx/(e^z+ze^z) - (e^y+ye^y)dy(e^z+ze^z)
u = f(x, y, z)
du = (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz
= (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)[(e^x+xe^x)dx/(e^z+ze^z) - (e^y+ye^y)dy(e^z+ze^z)]