Đáp án:
$2)\\ a)y'= 12 x^5- 15 x^4 - 8 x +6\\ b)y'=18x^2+2x−2\\ c) y'=-\dfrac{1}{\sqrt{x}(\sqrt{x}-1)^2}\\ d)y'=\dfrac{-1}{(x-1)^2}\\ e)y'=\dfrac{-6}{(2x-5)^2}\\ f)y'=\dfrac{x(x-2)}{(x-1)^2}\\ g)y'=\dfrac{4x^2+4x−14}{(2x+1)^2}\\ h)y'=1+\dfrac{2}{(x+1)^2}\\ i)y'=\dfrac{−5x^2+6x+8}{x^2+x+1}\\ k)y'=\dfrac{−2(x^2-1)}{(x^2-x+1)^2}$
Giải thích các bước giải:
$2)\\ a)y=(2x-3)(x^5-2x)\\ y'=(2x-3)'(x^5-2x)+(2x-3)(x^5-2x)'\\ =2(x^5-2x)+(2x-3)(5x^4-2)\\ = 12 x^5- 15 x^4 - 8 x +6\\ b)y=x(2x-1)(3x+2)\\ =(2x^2-x)(3x+2)\\ y'=(2x^2-x)'(3x+2)+(2x^2-x)(3x+2)'\\ =(4x-1)(3x+2)+3(2x^2-x)\\ =12 x^2+ 5 x-2 +6x^2−3x\\ =18x^2+2x−2\\ c)y=\left(\sqrt{x}+1\right)\left(\dfrac{1}{\sqrt{x}-1}\right)\\ y'=\left(\sqrt{x}+1\right)'\left(\dfrac{1}{\sqrt{x}-1}\right)+\left(\sqrt{x}+1\right)\left(\dfrac{1}{\sqrt{x}-1}\right)'\\ =\dfrac{1}{2\sqrt{x}}\left(\dfrac{1}{\sqrt{x}-1}\right)+\left(\sqrt{x}+1\right)\left((\sqrt{x}-1)^{-1}\right)'\\ =\dfrac{1}{2\sqrt{x}}\left(\dfrac{1}{\sqrt{x}-1}\right)-\left(\sqrt{x}+1\right)(\sqrt{x}-1)^{-2}(\sqrt{x}-1)'\\ =\dfrac{1}{2\sqrt{x}(\sqrt{x}-1)}-\left(\sqrt{x}+1\right)\dfrac{1}{(\sqrt{x}-1)^{2}}\dfrac{1}{2\sqrt{x}}\\ =\dfrac{1}{2\sqrt{x}(\sqrt{x}-1)}-\dfrac{\sqrt{x}+1}{2\sqrt{x}(\sqrt{x}-1)^{2}}\\ =\dfrac{\sqrt{x}-1-(\sqrt{x}+1)}{2\sqrt{x}(\sqrt{x}-1)^2}\\ =-\dfrac{1}{\sqrt{x}(\sqrt{x}-1)^2}\\ d)y=\dfrac{2x-1}{x-1}\\ y'=\dfrac{(2x-1)'(x-1)-(2x-1)(x-1)'}{(x-1)^2}\\ =\dfrac{2(x-1)-(2x-1)}{(x-1)^2}\\ =\dfrac{-1}{(x-1)^2}\\ e)y=\dfrac{3}{2x-5}\\ y'=\dfrac{3'(2x-5)-3(2x-5)'}{(2x-5)^2}\\ =\dfrac{-3.2}{(2x-5)^2}\\ =\dfrac{-6}{(2x-5)^2}\\ f)y=\dfrac{x^2+x-1}{x-1}\\ y'=\dfrac{(x^2+x-1)'(x-1)-(x^2+x-1)(x-1)'}{(x-1)^2}\\ =\dfrac{(2x+1)(x-1)-(x^2+x-1)}{(x-1)^2}\\ =\dfrac{x^2−2x}{(x-1)^2}\\ =\dfrac{x(x-2)}{(x-1)^2}\\ g)y=\dfrac{2x^2-4x+5}{2x+1}\\ y'=\dfrac{(2x^2-4x+5)'(2x+1)-(2x^2-4x+5)(2x+1)'}{(2x+1)^2}\\ =\dfrac{(4x-4)(2x+1)-2(2x^2-4x+5)}{(2x+1)^2}\\ =\dfrac{4x^2+4x−14}{(2x+1)^2}\\ =\dfrac{4x^2+4x−14}{(2x+1)^2}\\ h)y=x+1-\dfrac{2}{x+1}\\ y'=1+\dfrac{2}{(x+1)^2}\\ i)y=\dfrac{5x-3}{x^2+x+1}\\ y'=\dfrac{(5x-3)'(x^2+x+1)-(5x-3)(x^2+x+1)'}{x^2+x+1}\\ =\dfrac{5(x^2+x+1)-(5x-3)(2x+1)}{x^2+x+1}\\ =\dfrac{−5x^2+6x+8}{x^2+x+1}\\ k)y=\dfrac{x^2+x+1}{x^2-x+1}\\ y'=\dfrac{(x^2+x+1)'(x^2-x+1)-(x^2+x+1)(x^2-x+1)'}{(x^2-x+1)^2}\\ =\dfrac{(2x+1)(x^2-x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}\\ =\dfrac{−2x^2+2}{(x^2-x+1)^2}\\ =\dfrac{−2(x^2-1)}{(x^2-x+1)^2}$
