※ 引述《Applebao (寶兒)》之銘言:
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: 這題高階偏導數 找不出g和f的關係
: 也是沒看過的題型,麻煩大家幫忙了
f(x,y,z) = g(r) , r=(x^2+y^2+z^2)^(1/2)
=> @f/@x = g'(r) * @r/@x = g'(r)* (x/r)
=> @^2f/@x^2 = g"(r)*(x/r)^2 + g'(r)*@(x/r)/@x
= g"(r)*(x/r)^2 + g'(r)(1/r-x^2/r^3)
同理 @^2f/@y^2 = g"(r)*(y/r)^2 + g'(r)(1/r-y^2/r^3)
@^2f/@z^2 = g"(r)*(z/r)^2 + g'(r)(1/r-z^2/r^3)
所以 @^2f/@x^2 + @^2f/@y^2 + @^2f/@z^2
= g"(r)(x^2+y^2+z^2)/r^2 +g'(r)[3/r -(x^2+y^2+z^2)/r^3]
= g"(r) + g'(r)(3/r - 1/r)
= g"(r) + 2g'(r)/r
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