Question

求偏导数。

z=x/(x^2+y^2)^(1/2)的偏导数

Answer 1

z = x/√(x² + y²)
∂z/∂x = [√(x² + y²) - x · 1/(2√(x² + y²)) · 2x]/(x² + y²)
= (2x² + 2y² - 2x²)/(2(x² + y²)^(3/2))
= y²/(x² + y²)^(3/2)
∂z/∂y = x · - 1/(2(x² + y²)^(3/2)) · 2y
= - xy/(x² + y²)^(3/2)
= - yz/(x² + y²)