一道关于复合函数偏导数的求解问题,望高手解答,非常感谢。
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z = y/f(x² - y²),令u = x² - y²
∂z/∂x = y · [- f'(x² - y²) · 2x]/f(x² - y²)² = - 2xyf'(x² - y²)/f(x² - y²)²
∂z/∂y = [f(x² - y²) - y · f'(x² - y²) · - 2y]/f(x² - y²)² = [f(x² - y²) + 2y²f'(x² - y²)]/f(x² - y²)²
LHS = (1/x)∂z/∂x + (1/y)∂z/∂y
= (1/x) · [- 2xyf'(x² - y²)/f(x² - y²)²] + (1/y) · [f(x² - y²) + 2y²f'(x² - y²)]/f(x² - y²)²
= [- 2yf'(x² - y²) + (1/y)f(x² - y²) + 2yf'(x² - y²)]/f(x² - y²)²
= [(1/y)f(x² - y²)]/f(x² - y²)²
= 1/[yf(x² - y²)]
= 1/y² · y/f(x² - y²)
= 1/y² · z
= z/y²
= RHS
∂z/∂x = y · [- f'(x² - y²) · 2x]/f(x² - y²)² = - 2xyf'(x² - y²)/f(x² - y²)²
∂z/∂y = [f(x² - y²) - y · f'(x² - y²) · - 2y]/f(x² - y²)² = [f(x² - y²) + 2y²f'(x² - y²)]/f(x² - y²)²
LHS = (1/x)∂z/∂x + (1/y)∂z/∂y
= (1/x) · [- 2xyf'(x² - y²)/f(x² - y²)²] + (1/y) · [f(x² - y²) + 2y²f'(x² - y²)]/f(x² - y²)²
= [- 2yf'(x² - y²) + (1/y)f(x² - y²) + 2yf'(x² - y²)]/f(x² - y²)²
= [(1/y)f(x² - y²)]/f(x² - y²)²
= 1/[yf(x² - y²)]
= 1/y² · y/f(x² - y²)
= 1/y² · z
= z/y²
= RHS
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