设函数Z=(y-1)/f(x^2-y^2),其中f可导,试求z在(1,1)的导数dz

你好!“数学之美”团员448755083为你解答!
首先dz不叫导数,对于多元函数来讲,应该叫全微分.
∂f/∂x = f'·2x
∂f/∂y = -f'·2y
∂z/∂x = [ -(y - 1)·(∂f/∂x) ]/f² = [ -2x(y-1)f' ]/f²
∂z/∂y = [ f - (y - 1)·(∂f/∂y) ]/f² = [ f + 2y(y-1)f' ]/f²
∂z/∂x |(1,1) = 0
∂z/∂x |(1,1) = 1/f(0)
dz = (∂z/∂x)dx + (∂z/∂y)dy = dy/f(0)