1.设y=2x^3-12x^2+a在[-1,2]上最大值为2,求a 2.求由方程ycosx+sin（x-y）=0确定的隐函数y=f（x）的导数y
问题描述：
1.设y=2x^3-12x^2+a在[-1,2]上最大值为2,求a 2.求由方程ycosx+sin（x-y）=0确定的隐函数y=f（x）的导数y
答案是y'=[ysinx-cos(x-y)]/[cosx+cos(x-y)]
答
1、y=2x³-12x²+a
求导y'=6x²-24x
令y'=0 解得 x=0或x=4
∴y在x∈[-1,2]上先增后减,在x=0处取得最大值
∴y(x=0)=a=2
∴a=2
2、ycosx+sin（x-y）=0
求导得：y'cosx-ysinx+cos(x-y)*(1-y')=0
y'[cosx-cos(x-y)]=ysinx-cos(x-y)
y'=[ysinx-cos(x-y)]/[cosx-cos(x-y)]