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y 2xsin 2x 5 的导数

这是求导公式

y' =(2xsinx)' =2[x'sinx+x(sinx)'] =2(sinx+xcosx)

令f(x)=y=sin³2x f'(x)=3sin²2x·(sin2x)' =3sin²2x·cos2x·(2x)' =3sin²2x·cos2x·2 =3·(2sin2xcos2x)·sin2x =3sin4xsin2x =3·½·[cos(4x-2x)-cos(4x+2x)] =(3/2)cos2x -(3/2)cos6x 函数的导函数为(3/2)cos2x -(3/2)cos6x

-6sin^2(1-2x)cos(1-2x)

y=2xsin(2x-5) y'=[2xsin(2x-5)]' =2[xsin(2x-5)]' =2{sin(2x-5)+xcos(2x-5)*2} =2{sin(2x-5)+2xcos(2x-5)} ∫2xsin(2x-5)dx =∫xsin(2x-5)d(2x-5) =-∫xdcosx(2x-5) =-xcosx(2x-5)+∫cosx(2x-5)dx =-xcosx(2x-5)+(1/2)∫cosx(2x-5)d(2x-5) =-xcosx(2...

(sin²x)' = 2sinx(sinx)'  = 2sinxcosx  = sin2x 或: (sin²x)' = [(1-cos2x)/2]'  = [1/2 - (cos2x)/2]'  = 0 - ½(-sin2x)(2x)'  = ½(sin2x)×2  = sin2x

y'=6sin^22xcos2x=3sin4xsin2x

无法正常回答

因为y=sinx的倒数为y=cosx. 根据复合倒数的,此函数的倒数为y=2cos(2x+派除3) 看懂了吗?(不好意思,找不到那个符号)

复合函数求导,先把2x+5看成整体,把y(外部)求导,再把2x+5(内部)求导,内外相乘

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