Question

高数题，如图，求解？12题？

Answer 1

z=F(u)
F(u)= sinu +∫(1->2x-y) φ(t) dt ( 该是这样 )
To find: ∂z/∂x , ∂z/∂y

Solution:

2u= sinu +∫(1->2x-y) φ(t) dt
2u.∂u/∂x = cosu + φ(2x-y). (2∂u/∂x )
2( 1-φ(2x-y) ) .∂u/∂x = cosu
∂u/∂x = cosu/[2( 1-φ(2x-y) )] (1)

2u= sinu +∫(1->2x-y) φ(t) dt
2.∂u/∂y = cosy + φ(2x-y). (-∂u/∂y )
(2+φ(2x-y) ) .∂u/∂y = cosu
∂u/∂y = cosu/( 2+φ(2x-y) ) (2)

z=F(u)
∂z/∂x = Fx(u) . ∂u/∂x = Fx(u) . {cosu/[2( 1-φ(2x-y) )]}
∂z/∂y = Fy(u) . ∂u/∂y =Fy(u) .{ cosu/( 2+φ(2x-y) ) }

追问

谢谢