y=x^n,则导数y'=n*x^(n-1)y=f(x)+g(x),则导数y'=f'(x)+g'(x)所以x^3的导数=3*x^(3-1)=3x^2-2x^2导数=-2*[2*x^(2-1)]=-4x常数的导数是0，所以-1的导数=0所以y'=(x^3)'+(-2x^2)'+(-1)'=3x^2+(-4x)+0=3x^2-4xx=1,y'=3-4=-1所以y=x^3-2x^2-1在x=1处的导数为-1
