y=kx+b中的x怎么求(已知zf2xy)
y=kx+b中的x怎么求(已知zf2xy)
已知z=f(2xy x^2 y^2 x^3),且z对x y的所有二阶偏导数
一阶偏导数计算:z=f(2xy x^2 y^2 x^3),用全微分求导法,则有:
dz=2f1'(ydx xdy) f2'(2xdx 2ydy) 3x^2f3'dx,即:
dz=2yf1'dx 2xf1'dy 2xf2'dx 2yf2'dy 3x^2f3'dx,
dz=(2yf1' 2xf2' 3x^2f3')dx (2xf1' 2yf2')dy。
则z对x的一阶偏导数为:
∂z∂x =2yf1' 2xf2' 3x^2f3';
同理,z对y的一阶偏导数为:
∂z∂y =2xf1' 2yf2'。
二阶偏导数求解:因为∂z∂x =2yf1' 2xf2' 3x^2f3',再次对x求导,
所以∂^2z∂x^2
=2y(f11''*2y f12''*2x 3x^2f13'') 2f2' 2x(f21''2y f22''*2x 3x^2f23'') 6xf3' 3x^2(f31''2y f32''*2x 3x^2f33'')
=4y^2f11'' 8xyf12'' 6yx^2f13'' 2f2' 4x^2f22'' 6x^3f23'' 6xf3' 6yx^2f31'' 6x^3f32'' 9x^4f33''
=4y^2f11'' 8xyf12'' 6yx^2f13'' 2f2' 4x^2f22'' 12x^3f23'' 6xf3' 9x^4f33''
因为∂z∂y =2xf1' 2yf2',再次对y求导,
所以∂^2z∂y^2
=2x(f11''*2x f12''*2y f13''*0) 2f2' 2y(f21''*2x f22''*2y f23''*0)
=4x^2f11'' 4xyf12'' 2f2' 4xyf12'' 4y^2f22''
=4x^2f11'' 8xyf12'' 2f2' 4y^2f22''.
因为∂z∂y =2xf1' 2yf2',再次对x求导,
所以∂^2z∂y∂x
=2f1' 2x(f11''*2y f12''*2x 3x^2f13'') 2y(f21''*2y f22''*2x 3x^2f23'')
=2f1' 4xyf11'' 4x^2f12'' 6x^3f13'' 4y^2f12'' 4xyf22'' 6yx^2f23''
=2f1' 4xyf11'' 4(x^2 y^2)f12'' 6x^3f13'' 4xyf22'' 6yx^2f23''。