y=x+sinx
y'=1+cosx>=0,
因此y單調(diào)增,所以存在反函數(shù)
交換x,y,得反函數(shù)y=f^(-1)(x)滿足:
x=y+siny
對x求導(dǎo):1=y'+y'cosy,得;y'=1/(1+cosy)
x=1時,由1=y+siny,得:y=y0
因此y'(1)=1/(1+cosy0)就是解不出y0才問的……冒似出題應(yīng)該求的是y'(0),這樣的話y(0)=0, 則y'(0)=1/(1+cos0)=1/2如果求的是y'(1),得用數(shù)值解法求得y0=0.51097342938857...y'(0)=0.53411131613025...
證明f(x)=x+sinx一定存在反函數(shù)f^-1(x),并求(f^-1)'(1)
證明f(x)=x+sinx一定存在反函數(shù)f^-1(x),并求(f^-1)'(1)
數(shù)學(xué)人氣:644 ℃時間:2019-10-24 08:56:22
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