$1)y=2x^5-3x^4+4x^2+2x+1\\ y'=10x^4-12x^3+8x+2\\ 2)y=\dfrac{x+5}{-2x+3}\\ y'=\dfrac{(x+5)'(-2x+3)-(x+5)(-2x+3)'}{(-2x+3)^2}\\ =\dfrac{-2x+3+2(x+5)}{(-2x+3)^2}\\ =\dfrac{13}{(-2x+3)^2}\\ 3)y=(-x^2+x+1)\cos\sqrt{x+1}\\ y'=(-x^2+x+1)'\cos\sqrt{x+1}+(-x^2+x+1)\cos\sqrt{x+1}'\\ =(-2x+1)\cos\sqrt{x+1}-(-x^2+x+1)\sin\sqrt{x+1}.\sqrt{x+1}'\\ =(-2x+1)\cos\sqrt{x+1}+(x^2-x-1)\sin\sqrt{x+1}.\dfrac{1}{2\sqrt{x+1}}\\ 2\\ 1)y=-x^3+3x^2+9x-2\\ y'=-3x^2+6x+9\\ y' \le 0\\ \Leftrightarrow -3x^2+6x+9\le 0\\ \Leftrightarrow x^2-2x-3\ge 0\\ \Leftrightarrow (x-3)(x+1) \ge 0\\ \Leftrightarrow \left[\begin{array}{l} x\ge3\\ x\le-1\end{array} \right.$
2)Chụp lại đề
