1)
Đặt : [TEX]\Delta x=x-x_0[/TEX]
Ta có :[TEX] f(x_0+\Delta x) = (x_0+\Delta x)^3-3(x_0+\Delta x)^2+2(x_0+\Delta x)-1[/TEX]
[TEX]f(x_0)= x_0^3-3x_0^2+2x_0-1[/TEX]
=> [TEX] f(x_0+\Delta x)-f(x_0) = (x_0+\Delta x)^3-3(x_0+\Delta x)^2+2(x_0+\Delta x)-1 - (x_0^3-3x_0^2+2x_0-1) [/TEX]
[TEX]=\Delta x.[(\Delta x)^2+3x_0(x_0+\Delta x)-3\Delta x-6x_0+2]
[/TEX]
Ta có : $f'(x_0)=\lim_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$
$=\lim_{\Delta x\rightarrow 0}\frac{\Delta x.[(\Delta x)^2+3x_0(x_0+\Delta x)-3\Delta x-6x_0+2]}{\Delta x}$
$=\lim_{\Delta x\rightarrow 0} [(\Delta x)^2+3x_0(x_0+\Delta x)-3\Delta x-6x_0+2]=3x_0^2-6x_0+2$
Khi [TEX] x_0=1 [/TEX]
[TEX]f'(1) =3.1^2-6.1+2 =-1 [/TEX]
2)
Áp dụng [TEX](u.v)'= u'.v+u.v'[/TEX] thôi .