Câu hỏi:

$\underset{x\to 0}{lim } \frac{\sin 2x}{\tan 3x}$ bằng: 

A. 1

B. 0

C. 2/3

D. 3/2

Câu 1:

Tính đạo hàm của hàm số sau:$y={\left(2+{\sin }^{2}2x\right)}^3$ .

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

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

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

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

Câu 2:

Đạo hàm của hàm số $y={\cot }^2\left(\cos x\right)+\sqrt{\sin x-\frac{\pi }{2}}$ là

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

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

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

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

Câu 3:

Hàm số $y=2\sqrt{\sin x}-2\sqrt{\cos x}$ có đạo hàm là:

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

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

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

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

Câu 4:

Đạo hàm của hàm số $y=\sqrt{2+\tan \left(x+\frac{1}{x}\right)}$ là:

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

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

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

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

Câu 5:

Cho hàm số: 

$f\left(x\right)=\frac{\sin 3x}{3}-\cos x-\sqrt{3}\left(\sin x-\frac{\cos 3x}{3}\right)$. Giải phương trình f’(x)=0

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

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

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

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

Câu 6:

Tính đạo hàm của hàm số sau $y=\sin \left({\cos }^{2}x.{\tan }^{2}x\right)$

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

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

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

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