Tính đạo hàm?

[daocodai95]

\[(1-x)(1-x)^2(1-x^3)^3\]. tính đạo hàm:courage:

 

[tienk10]

đạo hàm: -(1-x)^2.(1-x^3)^3+(1-x).2.(-).(1-x).(1-x^3)^3 + (1-x).(1-x)^2.3.(-3x^2).(1-x^3)^2
hay -3(1-x)^2.(1-x^3)^3 - 9.x^2.(1-x).(1-x)^2.(1-X^3)^2
TA DÙNG CÔNG THỨC: U.V= U'.V+V'.U
Ở ĐÂY TA CÓ: U.V.Q = U'.V.Q+U.V'.Q +U.V.Q'

 

[letamnhi1995]

[FONT=&quot]bạn đạo hàm bình thường rồi giải u thôi! ^^!
y’ = (1-x)’(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]3[/SUP] + (1-x)[(1-x)[SUP]2[/SUP]]’(1-x[SUP]3[/SUP])[SUP]3[/SUP] + (1-x)(1-x)[SUP]2[/SUP][(1-x[SUP]3[/SUP])[SUP]3[/SUP]]’[/FONT]

[FONT=&quot]= -(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]3[/SUP] + (1-x)[2(1-x)(1-x)’](1-x[SUP]3[/SUP])[SUP]3[/SUP] + (1-x)(1-x)[SUP]2[/SUP][3(1-x[SUP]3[/SUP])[SUP]2[/SUP](1-x[SUP]3[/SUP])’][/FONT]

[FONT=&quot]= -(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]3 [/SUP]+ (1-x)[-2(1-x)](1-x[SUP]3[/SUP])[SUP]3[/SUP] + (1-x)(1-x)[SUP]2[/SUP][(3(1-x[SUP]3[/SUP])[SUP]2[/SUP] . -3x[SUP]2[/SUP]][/FONT]
[FONT=&quot]= -(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]3[/SUP] – 2(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]3[/SUP] -9x[SUP]2[/SUP](1-x)(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]2[/SUP][/FONT]
[FONT=&quot]=-3(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]3[/SUP] -9x[SUP]2[/SUP](1-x)(1-x)[SUP]2[/SUP](1-x[SUP]3[/SUP])[SUP]2[/SUP][/FONT]