更新1:
題目用錯了 是這個才對 X(X^2+ Y^2+ Z^2 )^((-1)/2)
偏微分求解
回答 (2)
2013-01-17 9:47 pm
✔ 最佳答案
x(x^2+y^2+z^2)^(－1/2) 對x偏微分Sol
w=y^2+z^2
∂x(x^2+w)^(－1/2)/ ∂x
=x*[∂(x^2+w)^(－1/2)/ ∂x]+(x^2+w)^(－1/2)
=x*[∂(x^2+w)^(－1/2)/ ∂(x^2+w)]*[∂(x^2+w)∂x]+(x^2+w)^(－1/2)
=x*(－1/2)(x^2+w)^(－3/2)*2x+(x^2+w)^(－1/2)
=－x^2*(x^2+w)^(－3/2)+(x^2+w)^(－1/2)
=－x^2/(x^2+w)^(3/2)+1/(x^2+w)^(1/2)
=(－x^2+x^2+w)/(x^2+W)^(3/2)
=w/(x^2+w)^(3/2)
=(y^2+z^2)/(x^2+y^2+z^2)^(3/2)
2013-01-17 9:00 am
X*(X^2* Y^2* Z^2 )^((-1)/2)
=X*[X^(-1)*Y^(-1)*Z^(-1)]
=Y^(-1)*Z^(-1)
= 1 / YZ
對X偏微分= 0 ？
=X*[X^(-1)*Y^(-1)*Z^(-1)]
=Y^(-1)*Z^(-1)
= 1 / YZ
對X偏微分= 0 ？
收錄日期: 2021-05-02 10:41:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130116000015KK06442