sin2x导数怎么求。求详细步骤
y\uff1dsin2x\u7684\u5bfc\u6570\u662f\uff1f\u600e\u4e48\u6c42y=sin2x
\uff1d2sinxcosx
\u6c42y\u7684\u5bfc\u6570
y\u2018\uff1d2[(sinx)'cosx+sinx(cosx)']
=2[(cosx)^2-(sinx)^2]
=2cos2x
y=sin2x
y\u2032=2cos2x
\u5148\u5bf9sin\u6c42\u5bfc\uff0c\u5f97\uff1acos2x
\u518d\u5bf92x\u6c42\u5bfc\uff0c\u5f97\uff1a2
\u7136\u540e\u76f8\u4e58\uff1ay\u2032=2cos2x
============
\u4e0d\u662f\u7684\uff0c\u4f60\u8bf4\u7684\u90a3\u4e2a\u516c\u5f0f\u662f\u4e24\u4e2a\u51fd\u6570\u76f8\u4e58\u65f6\uff0c\u6c42\u5bf9\u81ea\u53d8\u91cf\u7684\u5bfc\u6570\u3002
\u800cy=sin2x\uff0c\u662f\u590d\u5408\u51fd\u6570\u5bf9\u81ea\u53d8\u91cf\u6c42\u5bfc\u3002
令t=2x,所以sint的导数为cost
t的导数为2
所以(sin2x)'=2cos2x
(sin2x)′
=cos2x*(2x)′
=2cos2x
(sin2x)'=(cos2x)×(2x)'=2xcos2x 正解
(sin2x)'=(cos2x)×(2x)'=2cos2x
绛旓細鎵浠(sin2x)'=2cos2x
绛旓細sin2x鐨勫鏁锛2cos2x銆傞鍏堣浜嗚ВSinX鐨勫鏁版槸CosX銆傚啀鏍规嵁澶嶅悎鍑芥暟姹傚鍏紡Y'x=Y'u*Ux'銆傛妸2x鐪嬪仛涓涓暣浣搖銆傛眰sin2x鐨勫鏁帮紝灏辨槸鍏堟眰鍑簊inu鐨勫鏁般傜劧鍚庡啀鍦ㄥ2x姹傚銆傛渶鍚庣粨鏋滐細锛坰in2x锛'=(2x)'*(sinu)'=2cos2x銆
绛旓細sin2x=2sinxcosx銆傚垎鏋愯繃绋嬪涓嬶細sin(伪+尾)=sin伪cos尾+ sin尾cos伪 sin(伪-尾)=sin伪cos尾-sinB*cos伪 鏍规嵁sin(伪+尾)=sin伪cos尾+ sin尾cos伪鍙緱锛歴in2x=sin锛坸+x锛=sinxcosx+sinxcosx=2sinxcosx
绛旓細杩愮畻鏂规硶鏈変互涓嬩袱绉嶏細1.(sin²x)' = 2sinx(sinx)' = 2sinxcosx = sin2x銆2.(sin²x)' = [(1-cos2x)/2]' = [1/2 - (cos2x)/2]' = 0 - ½(-sin2x)(2x)' = ½(sin2x)脳2 = sin2x銆
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