Question

高等数学求偏导问题，求详细过程

Answer 1

g(x,y) = f[1/√(x^2+y^2)]
g'<x> = -[x/√(x^2+y^2)] f'[1/√(x^2+y^2)]
g'<y> = -[y/√(x^2+y^2)] f'[1/√(x^2+y^2)]
g''<xx> = -[y^2/√(x^2+y^2)^(3/2)] f'[1/√(x^2+y^2)]
+ [x^2/(x^2+y^2)] f''[1/√(x^2+y^2)]
g''<yy> = -[x^2/√(x^2+y^2)^(3/2)] f'[1/√(x^2+y^2)]
+ [y^2/(x^2+y^2)] f''[1/√(x^2+y^2)]
g''<xx> + g''<yy>
= -[1/√(x^2+y^2)] f'[1/√(x^2+y^2)] + f''[1/√(x^2+y^2)]

追问

g'(x)那里没看太懂

追答

g' = f'[1/√(x^2+y^2)] * ∂[1/√(x^2+y^2)]/∂x
= -[x/√(x^2+y^2)] f'[1/√(x^2+y^2)]

追问

那g''(x)呢？

追答

g'' = -[x/√(x^2+y^2)]' f'[1/√(x^2+y^2)]
-[x/√(x^2+y^2)] f''[1/√(x^2+y^2)] [1/√(x^2+y^2)]'
= -[y^2/√(x^2+y^2)^(3/2)] f'[1/√(x^2+y^2)]
+ [x^2/(x^2+y^2)] f''[1/√(x^2+y^2)]