Đáp án B
Ta có
Δ y = ( x 0 + Δ x ) 2 − ( x 0 + Δ x ) − ( x 0 2 − x 0 ) = △ x 2 + 2 x 0 Δ x − Δ x .
Nên
f ' ( x 0 ) = lim Δ x → 0 Δ y Δ x = lim Δ x → 0 ( Δ x ) 2 + 2 x 0 Δ x − Δ x Δ x = lim Δ x → 0 ( Δ x + 2 x 0 − 1 ) .
Vậy f ' ( x ) = lim Δ x → 0 ( Δ x + 2 x − 1 ) .