f(x)=sin(2x+π/3)
求導
f'(x)=2cos(2x+π/3)
=2cos(-2x-π/3)
=2sin[π/2-(-2x-π/3)]
=2sin(2x+5π/6)
=2sin[2(x+π/4)+π/3]
所以這個是由f(x)的圖像
向左平移π/4個單位,縱坐標變?yōu)樵瓉淼?倍
選 Cf(x)=sin(2x+π/3)的導函數(shù)為什么是f'(x)=2cos(2x+π/3)？而不是f'(x)=cos(2x+π/3)呢？是公式嗎？前面那個2是怎么來的？請指教，謝謝！f'(x)=cos(2x+π/3)這個是復合函數(shù)求導啊t=2xt'=2f'(x)=2cos(2x+π/3)我還是不理解，能展開來求一下嗎？我想看一下步驟，謝謝！f'(x)=[sin(2x+π/3)]'*[2x]'=cos(2x+π/3) * 2=2cos(2x+π/3)