(x^2+y^2)cos(1/(根号(x^2+y^2)))x^2+y^2不等于0,求偏导

Question

(x^2+y^2)cos(1/(根号(x^2+y^2)))x^2+y^2不等于0,求偏导

Answer 1

令z=(x²+y²)cos[1/√(x²+y²)]
那么
∂z/∂x
=∂(x²+y²)/∂x *cos[1/√(x²+y²)] +(x²+y²) *∂cos[1/√(x²+y²)] /∂x
显然
∂(x²+y²)/∂x=2x,
而∂cos[1/√(x²+y²)] /∂x= -sin[1/√(x²+y²)] *∂[1/√(x²+y²)]/∂x
又∂[1/√(x²+y²)]/∂x= -x /(x²+y²)^(3/2)
所以
∂z/∂x
=2x *cos[1/√(x²+y²)] + (x²+y²) *sin[1/√(x²+y²)] * x /(x²+y²)^(3/2)
=2x *cos[1/√(x²+y²)] + x/√(x²+y²) *sin[1/√(x²+y²)]
而x和y1是对称的
所以同样得到
∂z/∂y
=2y *cos[1/√(x²+y²)] + y/√(x²+y²) *sin[1/√(x²+y²)]