How do you find the derivative of sin^3(2x)?
Thanks for helping
How do you find the derivative of sin^3(2x)?
2011-06-17 11:46 am
回答 (4)
2011-06-17 11:54 am
✔ 最佳答案
chain rule3 sin^2 (2X) * 2 cos (2X) = 6 sin^2 (2X) cos (2X)
2011-06-17 12:19 pm
Put sin[2x] = p
So dp/dx = 2*cos[2x]
now d/dx { sin^3[2x] } = d/dp [p^3] * dp/dx --- chain rule.
= 3*p^2 * 2* cos[2x]
= 6*p^2*cos[2x] = 6*sin^2[2x]cos[2x]
So dp/dx = 2*cos[2x]
now d/dx { sin^3[2x] } = d/dp [p^3] * dp/dx --- chain rule.
= 3*p^2 * 2* cos[2x]
= 6*p^2*cos[2x] = 6*sin^2[2x]cos[2x]
參考: I L Maths
2011-06-17 12:12 pm
using chain rule and substitution method we can differentiate this
let y =sin^3(2x) and let u =2x
then y= sin^3u and u =2x
now dy/dx = dy/du * du/dx substitution differentiation]
dy/du = 3sin^2u cosu (chain rule)
du/dx =2
so now dy/dx = 3sin^2u cosu (2) = 2 (3) sin^2(2x) cos(2x) = 6 sin^2x cos2x
let y =sin^3(2x) and let u =2x
then y= sin^3u and u =2x
now dy/dx = dy/du * du/dx substitution differentiation]
dy/du = 3sin^2u cosu (chain rule)
du/dx =2
so now dy/dx = 3sin^2u cosu (2) = 2 (3) sin^2(2x) cos(2x) = 6 sin^2x cos2x
2011-06-17 12:00 pm
f (x) = [ sin (2x) ] ³
f ` (x) = 3 [ sin (2x) ] ² [ 2 sin x cos x ]
f ` (x) = 3 [ sin (2x) ] ² [ sin (2x) ]
f ` (x) = 3 sin ³ (2x)
f ` (x) = 3 [ sin (2x) ] ² [ 2 sin x cos x ]
f ` (x) = 3 [ sin (2x) ] ² [ sin (2x) ]
f ` (x) = 3 sin ³ (2x)
收錄日期: 2021-05-01 00:52:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110617034643AACvjvD