38 câu Dạng 2: Đạo hàm của hàm số lượng giác

$y^{\text{'}}=\frac{5+2{\tan }^{2}x}{2\sqrt{3x+2\tan x}}$.

$y^{\text{'}}=\frac{5-2{\tan }^{2}x}{2\sqrt{3x+2\tan x}}$

$y^{\text{'}}=\frac{-5+2{\tan }^{2}x}{2\sqrt{3x+2\tan x}}$

$y^{\text{'}}=\frac{-5-2{\tan }^{2}x}{2\sqrt{3x+2\tan x}}$

$\frac{1}{2}.$

$-\frac{1}{2}.$

– 1.

1.

$y^{\text{'}}=2x\cos x-x^2\sin x.$

$y^{\text{'}}=2x\cos x+x^2\sin x.$

$y^{\text{'}}=2x\sin x+x^2\cos x.$

$y^{\text{'}}=2x\sin x-x^2\cos x.$

$\cos x\cos \left(\cos x\right)+\sin x\sin \left(\sin x\right)$.

$-\left[\sin x\cos \left(\cos x\right)+\cos x\sin x\left(\sin x\right)\right]$

$-\left[\cos x\cos \left(\cos x\right)+\sin x\sin \left(\sin x\right)\right]$.

$\sin x\cos \left(\cos x\right)+\cos x\sin x\left(\sin x\right)$

$\sin 4x.$

$2-\sin 4x.$

$\cos 4x-\sin 4x.$

$-\sin 4x.$

– 30.

10.

– 1.

– 9.

$2\pi$

$\frac{8\pi }{3}.$.

$\frac{4\sqrt{3}}{3}.$

0.

0.

$\sqrt{2}.$

$\frac{\pi }{2}.$

$\frac{2\sqrt{2}}{\pi }.$

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

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

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

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

$\left|a\right|<\sqrt{5}.$

$\left|a\right|\ge \sqrt{5}.$

$\left|a\right|>5.$

$\left|a\right|<5.$

$y=1+\sin x$ .

$y=\cos x$

$y=1-\cos x$

$y=\sin x$ .

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

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

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

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

$f^{\text{'}}\left(\pi \right)=3{\sin }^2\left(a\pi \right).\cos \left(a\pi \right).$

$f^{\text{'}}\left(\pi \right)=0.$

$f^{\text{'}}\left(\pi \right)=3a{\sin }^2\left(a\pi \right).$

$f^{\text{'}}\left(\pi \right)=3a.{\sin }^2\left(a\pi \right).\cos \left(a\pi \right).$

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

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

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

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

$y^{\text{'}}\left(\frac{\pi }{6}\right)=1.$

$y^{\text{'}}\left(\frac{\pi }{6}\right)=-1.$

$y^{\text{'}}\left(\frac{\pi }{6}\right)=\sqrt{3}.$

$y^{\text{'}}\left(\frac{\pi }{6}\right)=-\sqrt{3}.$

6.

$2\sin 2x.$

0.

$2\cos 2x.$

$-\frac{\pi }{2}.$

$\frac{\pi \sqrt{3}}{2}.$

0.

$\frac{\pi }{2}.$

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

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

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

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

$y^{\text{'}}=\frac{1+{\cot }^{2}2x}{\sqrt{\cot 2x}}$.

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

$y^{\text{'}}=\frac{1+{\tan }^{2}2x}{\sqrt{\cot 2x}}$.

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

$y^{\text{'}}=\frac{1}{{\sin }^{2}2x}.$

$y^{\text{'}}=\frac{4}{{\cos }^{2}2x}.$

$y^{\text{'}}=\frac{4}{{\sin }^{2}2x}.$

$y^{\text{'}}=\frac{1}{{\cos }^{2}2x}.$

$y^{\text{'}}=\frac{\tan \frac{x}{2}}{{\cos }^2\frac{x}{2}}.$

$y^{\text{'}}=\frac{2\sin \frac{x}{2}}{{\cos }^2\frac{x}{2}}.$

$y^{\text{'}}=\frac{\sin \frac{x}{2}}{2{\cos }^3\frac{x}{2}}.$

$y^{\text{'}}={\tan }^3\frac{x}{2}.$

$y^{\text{'}}=\frac{\cos x-\sin x}{\cos x+\sin x}.$

$y^{\text{'}}=\frac{\cos x+\sin x}{\cos x-\sin x}.$

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

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

$y^{\text{'}}=\frac{3\sin 6x}{2\sqrt{\cos 6x}}$.

$y^{\text{'}}=\frac{-3\sin 6x}{\sqrt{\cos 6x}}$.

$y^{\text{'}}=\frac{3\sin 6x}{\sqrt{\cos 6x}}$.

$y^{\text{'}}=\frac{-3\sin 6x}{\sqrt{\cos 6x}}$

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

$\frac{2}{3}.$

$y^{\text{'}}=2x\tan x+\frac{x^2}{{\cos }^{2}x}+\frac{1}{2\sqrt{x}}.$

$y^{\text{'}}=2x\tan x+\frac{x^2}{{\cos }^{2}x}+\frac{1}{\sqrt{x}}.$

$\frac{\sqrt{3}}{2}.$

$\frac{\sqrt{2}}{2}.$

2.

1.

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

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

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

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

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

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

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

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