當y=(1-cosx)/sinx時 則y微分後 y'=?
回答 (2)
2015-10-20 1:43 pm
✔ 最佳答案
y=(1-cos x)/sin xy'=[(sin x)(sin x)-(1- cos x)(cos x)]/[(sin x)(sin x)]
y'=[(sin x)(sin x)- cos x+(cos x)(cos x)]/[(sin x)(sin x)]
y'=(1-cos x)/[(sin x)(sin x)]
2015-10-20 1:43 pm
Sol
y=(1－Cosx)/Sinx
ySinx=1－Cosx
yCosx+Sinxy’=Sinx
Sinxy’=Sinx－Cosx(1－Cosx)/Sinx
y’=1－Cosx(1－Cosx)/Sin^2 x
=1－Cosx/(1+Cosx)
=1/(1+Cosx)
y=(1－Cosx)/Sinx
ySinx=1－Cosx
yCosx+Sinxy’=Sinx
Sinxy’=Sinx－Cosx(1－Cosx)/Sinx
y’=1－Cosx(1－Cosx)/Sin^2 x
=1－Cosx/(1+Cosx)
=1/(1+Cosx)
收錄日期: 2021-04-18 13:56:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151020053457AAADsPb