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热心网友
简单分析一下,答案如图所示
热心网友
解:
f'x(x,y)
=[(3x²y-y³)(x²+y²)-(x³y-xy³)2x]/(x²+y²)²
=[(x^4)y+4x²y³-(y^5)]/(x²+y²)²
f'x(0,0)
=lim(Δx→0) [f(Δx,0)-f(0,0)]/Δx
=lim(Δx→0) 0/Δx
=0
f'y(x,y)
=[(x³-3xy²)(x²+y²)-(x³y-xy³)2y]/(x²+y²)²
=[(x^5)-4x³y²-x(y^4)]/(x²+y²)²
f'y(0,0)
=lim(Δy→0) [f(0,Δy)-f(0,0)]/Δy
=lim(Δy→0) 0/Δy
=0
f''xy(0,0)
=lim(Δy→0) [f'x(0,Δy)-f'x(0,0)]/Δy
=lim(Δy→0) -(Δy^5)]/Δy(Δy²)²
=-1
f''yx(0,0)
=lim(Δx→0) [f'x(Δx,0)-f'y(0,0)]/Δx
=lim(Δx→0) (Δx^5)]/Δx(Δx²)²
=1