u=x²+y²+z²,x=sin(yz),求∂u/∂y,∂u/∂z
发布网友
发布时间：2024-10-24 05:28
共2个回答
热心网友
时间：2024-11-05 00:46
∂z/∂x只对x求导数,而把y看作一个常数,∂z/∂x=(x+y)'sin(x-y)+(x+y)sin(x-y)'=
sin(x-y)+(x+y)cos(x-y)
∂z/∂y只对y求导数,而把x看作一个常数,∂z/∂y=(x+y)'sin(x-y)+(x+y)sin(x-y)'
=sin(x-y)+(x+y)[-cos(x-y)]
=sin(x-y)-(x+y)cos(x-y)
热心网友
时间：2024-11-05 00:50
x=sin(yz)
∂x/∂y = -zcos(yz)
∂x/∂z = -ycos(yz)
u=x^2+y^2+z^2
∂u/∂y
= 2x. ∂x/∂y + 2y
=-2xzcos(yz) +2y
∂u/∂z
= 2x. ∂x/∂z + 2z
=-2xycos(yz) +2z
