设Z=f(y/x,y),f有二阶连续偏导数,求az/ax,az/ay,az/axay,

设Z=f(y/x,y),f有二阶连续偏导数,求az/ax,az/ay,az/axay,
a2z/ax2,a2z/ay2,a2z/axay
lemon002 1年前 已收到2个回答 举报

evergreen_su 幼苗

共回答了18个问题采纳率:83.3% 举报

令u=y/x v=y
z=f(u,v)
az/ax=af/au*au/ax+af/av*av/ax=af/au*(-y/x^2)
az/ay=af/au*au/ay+af/av*av/ay=af/au*(1/x)+af/av
a^2z/axay=a(az/ax)/ay=af/au*(-1/x^2)+[a^2f/au^2*au/ay+a^2f/auav*av/ay]*(-y/x^2)
=af/au*(-1/x^2)+[a^2f/au^2*(1/x)+a^2f/auav]*(-y/x^2)
=af/au*(-1/x^2)+a^2f/au^2*(-y/x^3)+a^2f/auav*(-y/x^2)

1年前 追问

5

lemon002 举报

  a^2z/ay^2=?

举报 evergreen_su

a^2z/ax^2=a(az/ax)/ax=af/au*(2y/x^3)+a^2f/au^2*(y^2/x^4) a^2z/ay^2=a(az/ay)/ay=(a^2f/au^2*(1/x)+a^2f/auav)*(1/x)+a^2f/auav*(1/x)+a^2f/av^2 =a^2f/au^2*(1/x^2)+a^2f/auav*(2/x)+a^2f/av^2

俄地神呀 幼苗

共回答了315个问题 举报

f'(1)=af/a(y/x), f'(2)=af/ay
f''(1,1)=a^2f/[a(y/x)]^2=a[f'(1)]/a(y/x), f''(2,1)=a^2f/[aya(y/x)]=a[f'(2)]/a(y/x),
f''(1,2)=a^2[f/[a(y/x)ay]=a[f'(1)/ay, f''(2,2)=a^2f/[ay]^2 = a[f'(2)]/ay.

1年前

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