Find the derivatives
h(x) = sin(tan(cos(x)))
Many thanks !!
A. Maths Derivatives
2009-07-10 2:00 am
回答 (2)
2009-07-10 2:20 am
✔ 最佳答案
Make good use of chain rule to tackle the problem...h(x) = sin(tan(cosx))
h'(x) = d[sin(tan(cosx))]/dx
= d[sin(tan(cosx))]/d(tan(cosx)) X d(tan(cosx))/d(cosx) X d(cosx)/dx
= cos(tan(cosx)) X sec2(cosx) X (-sinx)
= -(sinx)cos(tan(cosx)) / cos2(cosx)
2009-07-09 18:20:27 補充：
Note that sec^2x = 1/cos^2x
參考: Physics king
2009-07-10 2:11 am
h(x) = sin(tan(cos(x)))
h'(x)=[cos(tan(cos(x)) ] [sec^2(cosx) ] [-sinx]
h'(x)=[cos(tan(cos(x)) ] [sec^2(cosx) ] [-sinx]
收錄日期: 2021-04-19 14:55:06
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