Giúp e câu 2 vs ạ đạo hàm

Bài 2. a) $y' = x^2 - 2x + 2$ b) $y' = 2x - \frac{2}{\sqrt{x}}$ c) $y' = \frac{-2(x-4) - (-2x+3)}{(x-4)^2} = \frac{-5}{(x-4)^2}$ d) $y' = \frac{2(x-1) - (2x+1)}{(x-1)^2} = \frac{-3}{(x-1)^2}$ e) $y' = \frac{(x^2+1) - (x-1)2x}{(x^2+1)^2} = \frac{1+x^2-2x^2+2x}{(x^2+1)^2} = \frac{1-x^2+2x}{(x^2+1)^2}$ f) $y' = \frac{(x-1)(2x-2) - (x^2-2x+3)}{(x-1)^2} = \frac{2x^2-4x+2 - x^2+2x-3}{(x-1)^2} = \frac{x^2-2x-1}{(x-1)^2}$ g) $y' = (\sqrt{x}+2)'(x^2+1) + (\sqrt{x}+2)(x^2+1)' = \frac{1}{2\sqrt{x}}(x^2+1) + (\sqrt{x}+2)2x = \frac{x^2+1}{2\sqrt{x}} + 2x\sqrt{x} + 4x$ h) $y' = ((x+1)^2)'(x^2-1) + (x+1)^2(x^2-1)' = 2(x+1)(x^2-1) + (x+1)^22x = 2(x+1)(x^2-1) + 2x(x+1)^2$ i) $y' = 10(2x^3+1)^9(2x^3+1)' = 10(2x^3+1)^96x^2 = 60x^2(2x^3+1)^9$ j) $y' = 20(1+x^2)^{19}(1+x^2)' = 20(1+x^2)^{19}2x = 40x(1+x^2)^{19}$ k) $y' = 2\ln(3x+2)\frac{1}{3x+2}3 = \frac{6\ln(3x+2)}{3x+2}$ l) $y' = \frac{1}{(x^2+1)\ln 2}2x = \frac{2x}{(x^2+1)\ln 2}$ m) $y' = 3\cos 3x$ n) $y' = 3\cos^2 2x(-\sin 2x)2 = -6\cos^2 2x\sin 2x$ o) $y' = 2\tan x \sec^2 x$ p) $y' = -\frac{1}{\sin^2(4-x^2)}(-2x) = \frac{2x}{\sin^2(4-x^2)}$ q) $y' = 2\cos(2x+\frac{\pi}{8})$ r) $y' = -4\sin(4x+\frac{\pi}{3})$ s) $y' = 2\sec^2(2x+\frac{\pi}{4})$ t) $y' = 2\sin(x+\frac{\pi}{4})\cos(x+\frac{\pi}{4}) = \sin(2x+\frac{\pi}{2}) = \cos 2x$ u) $y' = \sin^2 x + x2\sin x\cos x = \sin^2 x + 2x\sin x\cos x$ v) $y' = -2\sin x\cos x + 2\cos 2x = -\sin 2x + 2\cos 2x$ w) $y' = 3\cos 3x - 3\cos x$ x) $y' = \sec^2 x - \frac{1}{\sin^2 x}$ y) $y' = 3\sec^2(x+\frac{\pi}{4}) + 2\csc^2(\frac{\pi}{4}-x)$ z) $y' = -2\sin x - 2\sin(\frac{2\pi}{3}+x) + 2\sin(\frac{2\pi}{3}-x) = -2\sin x - 2\sin(\frac{2\pi}{3}+x) + 2\sin(\frac{2\pi}{3}-x) = -2\sin x - 2\left(\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x\right) + 2\left(\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x\right) = -2\sin x - \sqrt{3}\cos x - \sin x + \sqrt{3}\cos x - \sin x = -4\sin x$