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
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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)]
答案是y'=[ysinx-cos(x-y)]/[cosx+cos(x-y)]
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)]
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)]
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