求函数y=sinx的导数是多少,怎么推导

Question

最主要的是把过程写详细点啊,谢啦!!

Answer 1

lim(Δy/Δx)
Δx->0
=lim{[sin(x+Δx)-sin(x)]/Δx}
Δx->0
=lim[2cos(x+Δx/2)sin(Δx/2)/Δx]
Δx->0
=lim[cos(x+Δx/2)sin(Δx/2)/Δx/2]
Δx->0
由cos(x)的连续性，有limcos(x+Δx/2) = cos(x)
Δx->0
以及lim[sin(Δx/2)/Δx/2] = 1
Δx->0
故得
lim(Δy/Δx)
Δx->0
=limcos(x+Δx/2)*lim[sin(Δx/2)/Δx/2]
Δx->0 Δx->0
=cos(x)*1
=cos(x)

Answer 2

(sinx)'
=dsinx/dx
=[sin(x+dx)]/dx
=(sinxcosdx+sindxcosx)/dx
=[(sinx)*0+(dx)*cosx]/dx
=dx*cosx/dx
=cosx

Answer 3

y'=cosX

Answer 4

y=cosx
高中知识很难推导，记住就行了