Tính đạo hàm [tex]y= (x^{7} + 3x^{4} + 2)^{10}[/tex] [tex]y= (x-x^{2})^{32}[/tex] [tex]y = (3x -5x^{2}+4x^{3})^{10}(x+3)^{5}[/tex] [tex]y=(x^{2}-2x+3)(2x+1)^{3}[/tex]
$y=(3x-5x^2+4x^3)^{10}(x+3)^5$ $y'=[(3x-5x^2+4x^3)^{10}]'.(x+3)^5 + (3x-5x^2+4x^3)^{10}[(x+3)^5]' \\ =10(3x-5x^2+4x^3)^9.(3-10x+12x^2)+(3x-5x^2+4x^3)^{10}5x^4.1 \\ = \ ... \\ = 5x^9(x+3)^4(4x^2-5x+3)^9(28x^3+47x^2-51x+18)$ hoặc $y'= 5(x+3)^4[x(4x^2-5x+3)]^9(28x^3+47x^2-51x+18)$ $y=(x^2-2x+3)(2x+1)^3$ $y'=(x^2-2x+3)'.(2x+1)^3 + (x^2-2x+3)[(2x+1)^3]' \\ =(2x-2)(2x+1)^3+(x^2-2x+3).3(2x+1)^2.2 \\ =2(2x+1)^2[(x-1)(2x+1)+(x^2-2x+3).3] \\ = \ ... \\ = 2(2x+1)^2(5x^2-7x+8)$ hoặc $y'=40x^4-16x^3+18x^2+50x+16$
