Partial Differentiation
回答 (2)
2014-03-22 9:04 pm
✔ 最佳答案
if w=x^2+y^2-z^2 , x=ln t , y=t^3 , z=sin t^2w = (ln t)^2 + (t^3)^2 - [sin (t^2)]^2
dw/dt = [ (ln t)^2 ]' + [ (t^3)^2 ]' - { [sin (t^2)]^2 }'
dw/dt = 2 (ln t) (1/t) + 6t^5 - 2 sin(t^2) cos(t^2) ‧ 2t
dw/dt = (2/t) (ln t) + 6t^5 - 4t sin(t^2) cos(t^2)
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dw/dt = (2/t) (ln t) + 6t^5 - 2t sin(2t^2)
2014-03-24 4:58 am
收錄日期: 2021-04-13 20:19:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140322000051KK00059