Question

设Z=f(y/x,y),f有二阶连续偏导数,求az/ax,az/ay,az/axay，求高人解答，要过程或最后答案

a2z/ax2,a2z/ay2,a2z/axay

Answer 1

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.
ay/ax=0, a(y/x)/ax=-y/x^2

az/ax = af/ax = af/a(y/x) * a(y/x)/ax + af/ay * ay/ax = f'(1)* (-y/x^2) + f'(2)*0 = -yf'(1)/x^2,
az/ay = af/ay = af/a(y/x)* a(y/x)/ay + af/ay * ay/ay = f'(1)* (1/x) + f'(2)* (1) = f'(1)/x + f'(2),

a^2z/[axay] = a[az/ay]/ax = a[f'(1)/x + f'(2)]/ax = a[f'(1)/x]/ax + a[f'(2)]/ax
=a[f'(1)]/ax*(1/x) + a(1/x)/ax*f'(1) + a[f'(2)]/ax
=(1/x){a[f'(1)]/a(y/x)*a(y/x)/ax + a[f'(1)]/ay*ay/ax} + f'(1)*(-1/x^2) + a[f'(2)]/a(y/x)*a(y/x)/ax + a[f'(2)]/ay*ay/ax
=(1/x)[f''(1,1)*(-y/x^2) + 0] - f'(1)/x^2 + f''(2,1)*(-y/x^2) + 0
=-yf''(1,1)/x^3 - yf''(2,1)/x^2 - f'(1)/x^2

a^2z/[ax]^2=a[az/ax]/ax = a[-yf'(1)/x^2]/ax = a[f'(1)]/ax*(-y/x^2) + f'(1)*a(-y/x^2)/ax
=(-y/x^2){a[f'(1)]/a(y/x)*a(y/x)/ax + a[f'(1)]/ay*ay/ax} + f'(1)*(2y/x^3)
=(-y/x^2)[f''(1,1)*(-y/x^2) + 0] + f'(1)*2y/x^3
=y^2f''(1,1)/x^4 + 2yf'(1)/x^3

a^2z/[ay]^2=a[az/ay]/ay=a[f'(1)/x+f'(2)]/ay=(1/x)a[f'(1)]/ay+a[f'(2)]/ay
=(1/x){a[f'(1)]/a(y/x)*a(y/x)/ay + a[f'(1)]/ay*ay/ay} + a[f'(2)]/a(y/x)*a(y/x)/ay + a[f'(2)]/ay*ay/ay
=(1/x)[f''(1,1)*(1/x) + f''(1,2)] + f''(2,1)*(1/x) + f''(2,2)
=f''(1,1)/x^2 + [f''(1,2) + f'(2,1)]/x + f''(2,2)

Answer 2

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)

追问

　　a^2z/ay^2=?

追答

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