x^2+y^2 = 13
求d2y/d2x
(即d兩次)
中七微分題
2009-12-13 4:21 am
回答 (1)
2009-12-13 4:50 am
✔ 最佳答案
x^2 + y^2 = 13Differentiate wrt x
2x + 2y(dy/dx) =0 ... (1)
dy/dx = - 2x/2y = -x/y
Differentiate (1) wrt x
2 + 2(dy/dx)(dy/dx) + 2y(d^2y/dx^2) = 0
2 + 2(-x/y)^2 + 2y(d^2y/dx^2) = 0
yd^2y/dx^2 = - 1 - x^2/y^2
yd^2y/dx^2 = -(y^2 + x^2)/y^2
d^2y/dx^2 = -13/y^3 = +/-13/(x^2 - 13)^(3/2)
收錄日期: 2021-04-13 16:59:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091212000051KK01464