(sin²x)'
=lim[h→0] [sin²(x+h)-sin²x]/h
=lim[h→0] (sin(x+h)+sinx)(sin(x+h)-sinx)/h
和差化積
=lim[h→0] [2sin((x+h+x)/2)cos((x+h-x)/2)]*[2cos((x+h+x)/2)sin((x+h-x)/2)]/h
=lim[h→0] 4sin(x+h/2)cos(h/2)cos(x+h/2)sin(h/2)/h
分子中的sin(h/2)的等價(jià)無窮小是h/2,換成h/2后與分母的h消去,
=lim[h→0] 2sin(x+h/2)cos(h/2)cos(x+h/2)
=2sinxcosx
希望可以幫到你,如果解決了問題,請點(diǎn)下面的"選為滿意回答"按鈕,