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y=Cos1/x求导,求过程

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y=(-1)+1/[1-cos(x+y)] y'=0-[1-cos(x+y)]'/[1-cos(x+y)]^2 =-sin(x+y)*(1+y')/[1-cos(x+y)]^2 y'[1-cos(x+y)]^2=-sin(x+y)*(1+y') y'[1-cos(x+y)]^2+sin(x+y)y'=-sin(x+y) y'=-sin(x+y)/{[1-cos(x+y)]^2+sin(x+y)}

根据题意可把原函数转化为y=(secx)^2进行求导比较简单,答案为:y'=2secx*tanx

如下

设u=√(x^2-1) dy/dx=(dy/du)(du/dx)=-cosu[x/√(x^2-1)] =-xcos[√(x^2-1)]/[√(x^2-1)]

令y=1/(1+acosx) (a≠0),(cosx≠-1/a) y'=[1/(1+acosx) ]'=-(1+acosx)'/(1+acosx)^2=asinx/(1+acosx)^2

自外到内,链式法则。

y=cos1/x y'=-sin1/x*(1/x)' =-sin1/x*(-1/x^2) =1/x²sin1/x

y=cos x /(1-sin x) y1=[-sinx(1-sinx)-cosx(-cosx)]/(1-sinx)^2 =[-sinx+(sinx)^2+(cosx)^2]/(1-sinx)^2 =(1-sinx)/(1-sinx)^2 =1/(1-sinx)

分两步求导,先令cosx=t,求1/t3的导数然后再求cosx的导数。两个结果相乘。

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