23 câu Trắc nghiệm Đại số và Giải tích 11 Bài 3 (Có đáp án): Đạo hàm của các hàm số lượng giác

$\cos 2x-\sin \frac{x^2+1}{2}-\frac{1}{{\cos }^2\sqrt{x}}$

$2\cos 2x-x\sin \frac{x^2+1}{2}-\frac{1}{2\sqrt{x}{\cos }^2\sqrt{x}}$

$-2\cos x+x\sin \frac{x^2+1}{2}-\frac{1}{2\sqrt{x}{\cos }^2\sqrt{x}}$

$2\cos 2x+x\sin \frac{x^2+1}{2}-\frac{1}{2\sqrt{x}{\cos }^2\sqrt{x}}$

$\cos 2x.4{\cos }^{2}x+\frac{1}{{\sin }^2{\displaystyle \frac{1}{x^2}}}-\cos 2x4{\sin }^{3}x$

$2\cos 4x+\frac{2}{x^3{\sin }^2{\displaystyle \frac{1}{x^2}}}$

$2\cos 4x+\frac{2}{x{\sin }^2{\displaystyle \frac{1}{x^2}}}$

$2\cos 4x-\frac{2}{x^3{\sin }^2{\displaystyle \frac{1}{x^2}}}$

0

2

1/2

1/4

$\frac{\cos x+\sin x}{\sin x-\cos x}$

$\frac{-1}{{\cos }^2\left(x+{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}$

$\frac{1}{{\cos }^2\left(x+{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}$

$\frac{-2\sin x}{{\left(\sin x-\cos x\right)}^2}$

$\frac{3{\sin }^2\left(2x+1\right)}{2\sqrt{{\sin }^3\left(2x+1\right)}}$

$3\sqrt{s\mathrm{in}\left(2x+1\right)}\cos \left(2x+1\right)$

$\frac{3{\sin }^{2}2\left(2x+1\right)}{2\sqrt{{\sin }^3\left(2x+1\right)}}$

$3\sqrt{\sin \left(2x+1\right)}$

$\frac{15}{{\left(x-2\right)}^2}{\cos }^4\frac{x+1}{x-2}\sin \frac{x+1}{x-2}$

$-5{\cos }^4\frac{x+1}{x-2}\sin \frac{x+1}{x-2}$

$5{\cos }^4\frac{x+1}{x-2}\sin \frac{x+1}{x-2}$

$-\frac{15}{{\left(x-2\right)}^2}{\cos }^4\frac{x+1}{x-2}\sin \frac{x+1}{x-2}$

$\frac{1}{{\sin }^{2}x\sqrt{3+2\tan x}}$

$\frac{-1}{{\sin }^{2}x\sqrt{3+2\tan x}}$

$\frac{1}{{\cos }^{2}x\sqrt{3+2\tan x}}$

$\frac{-1}{{\cos }^{2}x\sqrt{3+2\tan x}}$

$\frac{2x-1}{{\sin }^2\left(x^2-x+1\right)}$

$-\frac{2x-1}{\sin \left(x^2-x+1\right)}$

$\frac{2x-1}{co{s}^2\left(x^2-x+1\right)}$

$\frac{1-2x}{{\sin }^2\left(x^2-x+1\right)}$

$-6{\cos }^{5}x\sin x$

$6{\cos }^{5}x\sin x$

$6{\sin }^{5}x\cos x$

$6{\cos }^5 x$

$x=\frac{\mathrm{\pi }}{12}+k\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\frac{-3\mathrm{\pi }}{8}+\mathrm{k}\frac{\mathrm{\pi }}{2} \left(\mathrm{k}\in \mathrm{\mathbb{Z}}\right)$

$x=\frac{-\mathrm{\pi }}{12}+k\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\frac{-3\mathrm{\pi }}{8}+\mathrm{k}\frac{\mathrm{\pi }}{2} \left(\mathrm{k}\in \mathrm{\mathbb{Z}}\right)$

$x=-\frac{\mathrm{\pi }}{12}+k\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\frac{3\mathrm{\pi }}{8}+\mathrm{k}\frac{\mathrm{\pi }}{2} \left(\mathrm{k}\in \mathrm{\mathbb{Z}}\right)$

$x=\frac{\mathrm{\pi }}{12}+k\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\frac{3\mathrm{\pi }}{8}+\mathrm{k}\frac{\mathrm{\pi }}{2} \left(\mathrm{k}\in \mathrm{\mathbb{Z}}\right)$

$\frac{-x^2.\sin 2x}{{(\cos x+x\sin x)}^2}$

$\frac{-x^2.{\sin }^{2}x}{{(\cos x+x\sin x)}^2}$

$\frac{-x^2.\cos 2x}{{(\cos x+x\sin x)}^2}$

${\left(\frac{x}{\cos x+x\sin x}\right)}^2$

$\pi +k2\pi$

$2\pi +k4\pi$

$2\pi +k\pi$

$\pi +k\pi$

$y\text{'}=\frac{1}{\sqrt{\sin x}}-\frac{1}{\sqrt{\cos x}}$

$y\text{'}=\frac{1}{\sqrt{\sin x}}+\frac{1}{\sqrt{\cos x}}$

$y\text{'}=\frac{\cos x}{\sqrt{\sin x}}+\frac{\sin x}{\sqrt{\cos x}}$

$y\text{'}=\frac{\cos x}{\sqrt{\sin x}}-\frac{\sin x}{\sqrt{\cos x}}$

$y\text{'}=\sin 8x+\frac{45}{2}cos5x.\sin 10x$

$y\text{'}=8\sin 8x+\frac{5}{2}cos5x.\sin 10x$

$y\text{'}=8\sin x+\frac{45}{2}cos5x.\sin 10x$

$y\text{'}=8\sin 8x+\frac{45}{2}cos5x.\sin 10x$

$y\text{'}=6\sin 4x{\left(2+{\sin }^{2}2x\right)}^3.$

$y\text{'}=3\sin 4x{\left(2+{\sin }^{2}2x\right)}^2.$

$y\text{'}=sin4x{\left(2+{\sin }^{2}2x\right)}^2.$

$y\text{'}=6\sin 4x{\left(2+{\sin }^{2}2x\right)}^2.$

$y^{\text{'}}=\frac{1}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\cdot$

$y^{\text{'}}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\cdot$

$y^{\text{'}}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}.\left(1-\frac{1}{x^2}\right).$

$y^{\text{'}}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}.\left(1+\frac{1}{x^2}\right).$

$y\text{'}=-2\cot \left(\cos x\right)\frac{1}{{\sin }^2\left(\cos x\right)}+\frac{\cos x}{2\sqrt{\sin x-\frac{\pi }{2}}}.$

$y\text{'}=2\cot \left(\cos x\right)\frac{1}{{\sin }^2\left(\cos x\right)}.\sin x+\frac{\cos x}{2\sqrt{\sin x-\frac{\pi }{2}}}.$

$y\text{'}=-2\cot \left(\cos x\right)\frac{1}{{\sin }^2\left(\cos x\right)}+\frac{\cos x}{\sqrt{\sin x-\frac{\pi }{2}}}.$

$y\text{'}=2\cot \left(\cos x\right)\frac{1}{{\sin }^2\left(\cos x\right)}.\sin x-\frac{\cos x}{\sqrt{\sin x-\frac{\pi }{2}}}.$

$-\frac{1+{\cos }^{2}x}{2{\sin }^{3}x}$

$-\frac{1+{\sin }^{2}x}{2{\sin }^{3}x}$

$\frac{1+{\sin }^{2}x}{2{\sin }^{3}x}$

$\frac{3{\sin }^{2}x}{{\left(1+\cos x\right)}^3}$

$\frac{{\sin }^{2}x}{{\left(1+\cos x\right)}^3}$

$\frac{3{\sin }^{2}x}{{\left(1+\cos x\right)}^2}$

$\frac{2{\sin }^{2}x}{{\left(1+\cos x\right)}^2}$

$\frac{3{\sin }^{2}x}{{\left(1+\cos x\right)}^3}$

$y\text{'}=\cos \left({\cos }^{2}x.{\tan }^{2}x\right)\left(-\sin 2x{\tan }^{2}x+2\tan x\right)$

$y\text{'}=\cos \left({\cos }^{2}x.{\tan }^{2}x\right)\left(\sin 2x{\tan }^{2}x+\tan x\right)$

$y\text{'}=\cos \left({\cos }^{2}x.{\tan }^{2}x\right)\left(-\sin 2x{\tan }^{2}x+\tan x\right)$

$y\text{'}=\cos \left({\cos }^{2}x.{\tan }^{2}x\right)\left(\sin 2x{\tan }^{2}x+2\tan x\right)$

$\frac{6}{{\left(2\sin 2x-\cos 2x\right)}^2}$

$\frac{-6}{{\left(\sin 2x-\cos 2x\right)}^2}$

$\frac{6}{{\left(2\sin 2x-\cos x\right)}^2}$

$\frac{-6}{{\left(2\sin 2x-\cos 2x\right)}^2}$

$y\text{'}=\sin \left(2\cos \left({\tan }^{4}3x\right)\right).\left(\sin \left({\tan }^{4}3x\right)\right).4{\tan }^{3}3x.\left(1+{\tan }^{2}3x\right).3$

$y\text{'}=\sin \left(2\cos \left({\tan }^{4}3x\right)\right).\left(\sin \left({\tan }^{4}3x\right)\right).{\tan }^{3}3x.\left(1+{\tan }^{3}3x\right).$

$y\text{'}=-\sin \left(2\cos \left({\tan }^{4}3x\right)\right).\left(\sin \left({\tan }^{4}3x\right)\right).4{\tan }^{3}3x.\left(1+{\tan }^{2}3x\right).3$

$y\text{'}=\sin \left(2\cos \left({\tan }^{4}3x\right)\right).\left(\sin \left({\tan }^{4}3x\right)\right).4{\tan }^{3}3x$