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

Ta có: $y^{\text{'}}={\left(\sin x\right)}^{\text{'}}.\cos 4x+\sin x.{\left(\cos 4x\right)}^{\text{'}}$

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

Ta có $y^{\text{'}}=-3\sin \left(3x-\frac{\pi }{6}\right)+2\cos \left(\frac{2\pi }{3}-2x\right)$

$\Rightarrow y^{\text{'}}\left(\frac{\pi }{3}\right)=-3\sin \frac{5\pi }{6}+2\cos 0=\frac{1}{2}.$

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

Ta có: $y^{\text{'}}={\left[\cot \left(3x^2-5\right)\right]}^{\text{'}}=-\frac{6x}{{\sin }^2\left(3x^2-5\right)}$.

Ta có: $f^{\text{'}}\left(x\right)=\frac{{\left(\tan x+\cot x\right)}^{\text{'}}}{2\sqrt{\tan x+\cot x}}$

$=\frac{\frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}{2\sqrt{\tan x+\cot x}}$.

$=\frac{{\sin }^{2}x-{\cos }^{2}x}{2{\sin }^{2}x{\cos }^{2}x\sqrt{\tan x+\cot x}}$

$=\frac{-2\cos 2x}{{\sin }^{2}2x\sqrt{\tan x+\cot x}}$

Suy ra $f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{-2\cos \frac{\pi }{2}}{{\sin }^2\left(\frac{\pi }{2}\right)\sqrt{\tan \frac{\pi }{4}+\cot \frac{\pi }{4}}}=0$.

Ta có $y=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos x}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{{\cos }^2\frac{x}{2}}}}$

$=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{2}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{{\cos }^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{4}}$

$=\sqrt{{\cos }^2\frac{x}{8}}=\cos \frac{x}{8}.$

Do đó $y^{\text{'}}={\left(\cos \frac{x}{8}\right)}^{\text{'}}=-\frac{1}{8}\sin \frac{x}{8}$ .

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

Ta có:

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

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

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

Ta có: $VT=y^{\text{'}}{\left(\sin x-x\cos x\right)}^2-x^2{y}^2$

$=\frac{x^2}{{\left(\cos x+x\sin x\right)}^2}.{\left(\sin x-x\cos x\right)}^2-x^2.{\left(\frac{\sin x-x\cos x}{\cos x+x\sin x}\right)}^2=0=VP.$

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

Đáp án B

Bước đầu tiên sử dụng đạo hàm hàm tổng, sau đó sử dụng ${\left(\sin u\right)}^{\text{'}},{\left(\cos u\right)}^{\text{'}}$

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

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

Đáp án D

Ta có $y={\sin }^{4}x+{\cos }^{4}x=1-\frac{1}{2}{\sin }^{2}2x=\frac{3}{4}+\frac{1}{4}\cos 4x$

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

Đáp án B

Ta có: $y^{\text{'}}=10\cos 2x+20\sin 5x$ , suy ra $\left\{\begin{array}{l}a=20\\ b=10\end{array}\right.$. Vậy $a-b=10$ .

Đáp án D

Ta có $f^{\text{'}}\left(x\right)=\frac{2}{\cos \left(\pi x\right)}=2.{\left(\cos \left(\pi x\right)\right)}^{\text{'}}.\frac{-1}{{\cos }^2\left(\pi x\right)}=2.\pi \frac{\sin \left(\pi x\right)}{{\cos }^2\left(\pi x\right)}\Rightarrow f^{\text{'}}\left(3\right)=2\pi .\frac{\sin 3\pi }{{\cos }^{2}3\pi }=0$

Đáp án A

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

Đáp án B

Ta có $f^{\text{'}}\left(x\right)=2\cos x-a\sin x-3=0$ có nghiệm khi và chỉ khi $4+a^2\ge 9\iff a^2\ge 5\iff \left|a\right|\ge \sqrt{5}$

Đáp án D

Ta có $y^{\text{'}}=\cos x\Rightarrow y=\sin x+C$ (C: hằng số).

$f\left(\frac{\pi }{2}\right)=1\iff \sin \frac{\pi }{2}+C=1\iff C=0$. Vậy $y=\sin x$

Đáp án D

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

Đáp án B

Ta có $f\left(x\right)={\sin }^{3}ax\Rightarrow f^{\text{'}}\left(x\right)=3a{\sin }^{2}ax\cos ax\Rightarrow f^{\text{'}}\left(\pi \right)=3a{\sin }^{2}a\pi .\cos a\pi =0$

Đáp án A

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

Đáp án D

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

Suy ra $y^{\text{'}}\left(\frac{\pi }{6}\right)=\frac{-2.\frac{\sqrt{3}}{2}\left(1-\frac{1}{2}\right)+\frac{1}{2}.\frac{\sqrt{3}}{2}}{{\left(1-\frac{1}{2}\right)}^2}=\frac{-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}}{\frac{1}{4}}=4\left(-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}\right)=-2\sqrt{3}+\sqrt{3}=-\sqrt{3}$

Đáp án C

Ta có: $f\left(x\right)=\frac{1+\cos \left(\frac{2\pi }{3}-2x\right)}{2}+\frac{1+\cos \left(\frac{2\pi }{3}+2x\right)}{2}+\frac{1+\cos \left(\frac{4\pi }{3}-2x\right)}{2}+\frac{1+\cos \left(\frac{4\pi }{3}+2x\right)}{2}-2{\sin }^{2}x$

$\iff f\left(x\right)=2+\frac{1}{2}\left[\cos \left(\frac{2\pi }{3}-2x\right)+\cos \left(\frac{4\pi }{3}-2x\right)\right]+\frac{1}{2}\left[\cos \left(\frac{2\pi }{3}+2x\right)+\cos \left(\frac{4\pi }{3}+2x\right)\right]-2{\sin }^{2}x$$\iff f\left(x\right)=2+\cos \left(\pi -2x\right).\cos \frac{\pi }{3}+\cos \left(\pi +2x\right).\cos \frac{\pi }{3}-2{\sin }^{2}x$.

$\iff f\left(x\right)=2+\frac{1}{2}\left[\cos \left(\pi -2x\right)+\cos \left(\pi +2x\right)\right]-2{\sin }^{2}x$

$\iff f\left(x\right)=2-\cos 2x-2{\sin }^{2}x\iff f\left(x\right)=1$

Suy ra $f^{\text{'}}\left(x\right)=0$

Đáp án C

Ta có $y^{\text{'}}=\cos \left(\pi \sin x\right).{\left(\pi .\sin x\right)}^{\text{'}}=\pi \cos x.\cos \left(\pi \sin x\right)$

Suy ra $f^{\text{'}}\left(\frac{\pi }{6}\right)=\pi \cos \frac{\pi }{6}.\cos \left(\pi \sin \frac{\pi }{6}\right)=\pi .\frac{\sqrt{3}}{2}.\cos \left(\frac{\pi }{2}\right)=0.$

Đáp án B

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

Đáp án C

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

Đáp án A

Ta có $y^{\text{'}}=2\tan \frac{x}{2}.{\left(\tan \frac{x}{2}\right)}^{\text{'}}=\frac{\tan \frac{x}{2}}{{\cos }^2\frac{x}{2}}$

Đáp án C

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

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

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

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

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

Đáp án D

Ta có $y^{\text{'}}={\left(\sqrt{\cos 6x}\right)}^{\text{'}}=\frac{{\left(\cos 6x\right)}^{\text{'}}}{2\sqrt{\cos 6x}}=\frac{-6\sin 6x}{2\sqrt{\cos 6x}}=\frac{-3\sin 6x}{\sqrt{\cos 6x}}$

Đáp án C

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

Đáp án D

Ta có $f\left(\sin x+1\right)+f\left(\cos x\right)={\cos }^2\left(x-\frac{\pi }{4}\right)$ , đạo hàm 2 vế ta được

$\cos x.f^{\text{'}}\left(\sin x+1\right)-\sin x{f}^{\text{'}}\left(\cos x\right)=-2\cos \left(x-\frac{\pi }{4}\right).\sin \left(x-\frac{\pi }{4}\right)$

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

Thay$x=0$  vào phương trình (*), ta được $f^{\text{'}}\left(1\right)=-\sin \left(-\frac{\pi }{2}\right)\iff f^{\text{'}}\left(1\right)=1$

Đáp án C

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

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

Đáp án C

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