z=(x∧2+y∧2)sin(1/√(x∧2+y∧2))求对x,y的偏导
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/11/08 09:10:09
z=(x∧2+y∧2)sin(1/√(x∧2+y∧2))求对x,y的偏导
令1/√(x^2+y^2)=u,则z=1/u^2*sinu
dz/dx=dz/du*du/dx
=(-2/u^3*sinu+1/u^2*cosu)*[-x(x^2+y^2)^(-3/2)]
={2(x^2+y^2)^(3/2)*sin[1/√(x^2+y^2)]-(x^2+y^2)*cos[1/√(x^2+y^2)]}*[x(x^2+y^2)^(-3/2)]
=2xsin[1/√(x^2+y^2)]-x/√(x^2+y^2)*cos[1/√(x^2+y^2)]
dz/dy=dz/du*du/dy
=2ysin[1/√(x^2+y^2)]-y/√(x^2+y^2)*cos[1/√(x^2+y^2)]
dz/dx=dz/du*du/dx
=(-2/u^3*sinu+1/u^2*cosu)*[-x(x^2+y^2)^(-3/2)]
={2(x^2+y^2)^(3/2)*sin[1/√(x^2+y^2)]-(x^2+y^2)*cos[1/√(x^2+y^2)]}*[x(x^2+y^2)^(-3/2)]
=2xsin[1/√(x^2+y^2)]-x/√(x^2+y^2)*cos[1/√(x^2+y^2)]
dz/dy=dz/du*du/dy
=2ysin[1/√(x^2+y^2)]-y/√(x^2+y^2)*cos[1/√(x^2+y^2)]
z=(x∧2+y∧2)sin(1/√(x∧2+y∧2))求对x,y的偏导
设函数z=z(x,y)由方程2sin(x+2y-3z)=x+2y-3z所确定,求证z对x的偏导加上z对y的偏导等于1
已知(z-x)∧2-4(x-y)(y-x)=0求x+z-2y+8的值
三角函数最值问题已知x,y,z为实数,求:f(x,y,z)=[sin(x-y)]^2+[sin(y-z)]^2+[sin
z=y/(√(x^2+y^2)) -cosx 求对x的偏导.z等于根号x平方加y平方分之y,再减cos x,分别求对x,
x+y+z+sin(x+2y+3z)=0,求Zx偏导.
设z=z(x,y)是方程x^2+z^2=ysin(z/x)确定的隐函数,求Z对x,y的偏导数
证明sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((x+y)/2)sin((x+
已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值
根号x+根号y-1+根号z-2=1/2(x+y+z),求x,y,z的值
已知{x:y:z=1:2:3,x+y+z=12,求x、y、z的值
已知x::y:z=3:4:5,(1)求x+y分之z的值;(2)若x+y+z=6,求x,y,z.