昨日答案

设\(z=e^{xy}+f\left( x+y,xy\right) ,\)求\(\dfrac{\partial ^{2}z}{\partial x\partial y}\),其中$f\left( u,v\right) $有二阶连续偏导数.

解：z = eˣʸ + f(x + y , xy)

zₓ = y·eˣʸ + f₁’ + y·f₂’

az/axay = eˣʸ + xy·eˣʸ + f₁₁’’ + x·f₁₂’’ + f₂’ + y·f₂₁’’ + xy·f₂₂’’ = eˣʸ(1 + xy) + f₁₁’’ + (x + y)f₁₂’’ + f₂’ + xy·f₂₂’’

二阶偏导数连续→二阶偏导数相等

即 az/axay = az/ayax

每日一题