y′=(2x+1)′(x–3)–(x–3)′(2x+1)(x–3)2=–7(x–3)2 21/08/2021 Bởi Serenity y′=(2x+1)′(x–3)–(x–3)′(2x+1)(x–3)2=–7(x–3)2
$y’=(2x+1)'( x-3)-(x-3)'(2x+1).(x-3)^2$ $=2(x-3)-(2x+1)(x-3)$ $=(x-3)(2-2x-1)$ $=(x-3)(-2x+1)$ $=-2x^2+7 x-3$ Bình luận
Đáp án: $y′=(2x+1)′(x–3)–(x–3)′(2x+1)$ $=(2x+1)'(x-3)-(2x+1)(x-3)’$ $= 2(x-3)-(2x+1).1$ $= 2x-6-2x-2$ $= -8$ BẠN THAM KHẢO. Bình luận
$y’=(2x+1)'( x-3)-(x-3)'(2x+1).(x-3)^2$
$=2(x-3)-(2x+1)(x-3)$
$=(x-3)(2-2x-1)$
$=(x-3)(-2x+1)$
$=-2x^2+7 x-3$
Đáp án:
$y′=(2x+1)′(x–3)–(x–3)′(2x+1)$
$=(2x+1)'(x-3)-(2x+1)(x-3)’$
$= 2(x-3)-(2x+1).1$
$= 2x-6-2x-2$
$= -8$
BẠN THAM KHẢO.