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求函数y=sinnx的导数是多少

实际上再进行几步的化简即可 y=sinnx *(sinx)^n 那么求导得到 y'=(sinnx)' *(sinx)^n +sinnx *[(sinx)^n]' =n *cosnx *(sinx)^n +sinnx * n *(sinx)^n-1 *cosx 提取出n *(sinx)^n-1 *cosx 得到y'=n *(sinx)^n-1 *(sinx *cosnx +sinnx *cosx) 而...

d[sin(nx)]=cos(nx)d(nx)=ncosnxdx d[sin(nx)]/dx=ncosnx

这涉及到复合函数求导问题,要逐层求导,先对sin nx求导,再对nx求导

y'=n(sinx)^(n-1) cosx sin(nx)+n(sinx)^n* cos(nx)

[sin^n(x)]'=nsin^(n-1)(x)cosx [cosnx]'=-nsinnx y'=[sin^n(x)]'cos nx +[cosnx]'sin^n(x) =nsin^(n-1)(x)cosxcos nx-nsinnxsin^n(x) =nsin^(n-1)(x)(coscosnx-sinnxsinx) =nsin^(n-1)(x)cos(n+1)x cos(a+b)=cosacosb-sinasinb 公式啊,哥哥

y'={[(sinx)^n][(cosx)^n]}' =ncosx(sinx)^(n-1)-nsinx(cosx)^(n-1)

lny=nlnsinx+lncosnx1/y*y'=ncosx/sinx-nsinnx/cosnxy'=(ncosx/sinx-nsinnx/cosnx)*(sin^n*cosnx)

y′=nsinn-1xcosxcosnx-nsinnxsinnx=nsinn-1x(cosxcosnx-sinxsinnx)=nsinn-1xcos(n+1)x.故选D.

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