z=(x∧2+y∧2)sin(1/√(x∧2+y∧2))求对x,y的偏导
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z=(x∧2+y∧2)sin(1/√(x∧2+y∧2))求对x,y的偏导
z=(x∧2+y∧2)sin(1/√(x∧2+y∧2))求对x,y的偏导
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)]
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