设函数f(x)满足f'(lnx)=1-x,且f(0)=0,求f(x)
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![设函数f(x)满足f'(lnx)=1-x,且f(0)=0,求f(x)](/uploads/image/z/11567120-32-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%27%28lnx%29%3D1-x%2C%E4%B8%94f%280%29%3D0%2C%E6%B1%82f%28x%29)
设函数f(x)满足f'(lnx)=1-x,且f(0)=0,求f(x)
设函数f(x)满足f'(lnx)=1-x,且f(0)=0,求f(x)
设函数f(x)满足f'(lnx)=1-x,且f(0)=0,求f(x)
设Y=lnX,则X=e^y
所以f^(y)=1-e^y,即f(y)=y-e^y+a,a为常数
又因为f(0)=0,所以f(0)=0-1+a=0,所以a=1
所以f(x)=x-e^x=1,再加上一个x的范围负无穷到正无穷.