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z=y/f(x^2+y^2)的偏导数,分别对x、y求偏导
人气:330 ℃ 时间:2020-06-06 08:08:25
解答
z = y/f(x² + y²),令u = x² + y²
∂z/∂x = y · - 1 · [∂f(u)/∂u · ∂(x² + y²)/∂x]/[f(u)]²
= - y · f'(u) · 2x/[f(u)]²
= - 2xyf'(u)/[f(u)]²
= - 2xyf'(x² + y²)/[f(x² + y²)]²
∂z/∂y = [f(u) · ∂y/∂y - y · ∂f(u)/∂u · ∂(x² + y²)/∂y]/[f(u)]²
= [f(u) - yf'(u) · 2y]/[f(u)]²
= 1/f(u) - 2y²f'(u)/[f(u)]²
= 1/f(x² + y²) - 2y²f'(x² + y²)/[f(x² + y²)]²
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