f(x)=x+sinx
f′=1+cosx ≥0 ,f′=0,x=2kπ+π 導數(shù)為零的點是離散點,所以f(x)是單調遞增的,
而f′=1+cosx 顯然不是單調函數(shù).
所以單調函數(shù)的導數(shù)不一定是單調函數(shù).
當然也有單調函數(shù)的導數(shù)是單調函數(shù)的.例如
x＞0,f(x)=x²,是單調函數(shù),它的導數(shù)2x也是單調函數(shù).