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

$\begin{array}{l}\underset{x\to 0}{\lim }\frac{\sin 2x}{\tan 3x}=\underset{x\to 0}{\lim }\frac{\sin 2x.c\text{os 3x}}{\sin 3x}\\ =\underset{x\to 0}{\lim }\frac{2.\frac{\sin 2x}{2x}.c\text{os3x}}{3.\frac{\sin 3x}{3x}}=\frac{\mathrm{2.1.}c\text{os0}}{3.1}=\frac{2}{3}\end{array}$

ĐÁP ÁN C

$y\text{'}=c\text{os2x. (2x)' - sin}\frac{x{}^2+1}{2}.\left(\frac{x{}^2+1}{2}\right)\text{'}-\frac{1}{{\text{cos}}^2\sqrt{x}}.\left(\sqrt{x}\right)\text{'}$

Chọn B

Chú ý. Rút gọn trước khi tính đạo hàm

Chọn C

Chọn B

Chọn B

Chọn A

Chọn C

Chọn D

Chọn A

y = cos6 x+ sin2xcos2x(sin2x + cos2x) + sin4x - sin2x

= cos6x + sin2x(1 - sin2x).1 + sin4x - sin2x

 =  cos6x + sin2x - sin4x + sin4x - sin2x

= cos6x

Do đó : y' = -6cos5xsinx.

Chọn C

Chọn D

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left(\mathrm{s}\text{in}x-x\cos x\right)}^{\text{'}}\left(\cos x+x\sin x\right)-{\left(\cos x+x\sin x\right)}^{\text{'}}\left(\mathrm{s}\text{in}x-x\cos x\right)}{{\left(\cos x+x\sin x\right)}^2}\\ =\frac{\left(\text{cos}x-\cos x+\text{}x.\mathrm{s}\text{inx}\right)\left(\cos x+x\sin x\right)-\left(-\sin x+\sin x+x.c\text{osx}\right)\left(\mathrm{s}\text{in}x-x\cos x\right)}{{\left(\cos x+x\sin x\right)}^2}\\ =\frac{x\sin x\left(\cos x+x\sin x\right)-x\cos x\left(\mathrm{s}\text{in}x-x\cos x\right)}{{\left(\cos x+x\sin x\right)}^2}\\ =\frac{x^2{\sin }^{2}x+\text{}{x}^2{\cos }^{2}x}{{(\cos x+x\sin x)}^2}={\left(\frac{x}{\cos x+x\sin x}\right)}^2\end{array}$

Chọn B.

$\begin{array}{l}{y}^{\text{'}}={\left({\cot }^2\frac{x}{4}\right)}^{\text{'}}\\ =2\cot \frac{x}{4}{\left(\cot \frac{x}{4}\right)}^{\text{'}}\\ =-\frac{1}{2}\cot \frac{x}{4}\left(1+{\cot }^2\frac{x}{4}\right)\end{array}$

$\begin{array}{l}y\text{'}=0\\ \iff \frac{1}{2}\cot \frac{x}{4}\left(1+{\cot }^2\frac{x}{4}\right)=0\end{array}$

$\begin{array}{l}\iff \cot \frac{x}{4}=0\\ \iff \frac{x}{4}=\frac{\pi }{2}+k\pi \\ \iff x=2\pi +k4\pi ,k\in \mathbb{Z}\end{array}$

Chọn C

$\begin{array}{l}y\text{'}=2\left(\sqrt{\sin x}\right)\text{'}-2\left(\sqrt{\cos x}\right)\text{'}\\ =2.\cos x.\frac{1}{2\sqrt{\sin x}}+2\sin x\frac{1}{2\sqrt{\cos x}}\end{array}$

$=\frac{\cos x}{\sqrt{\sin x}}+\frac{\sin x}{\sqrt{\cos x}}$

Chọn D.

Bước đầu tiên áp dụng ${\left(u+v\right)}^{/}$ 

$y\text{'}={\left(2{\sin }^{2}4x\right)}^{/}-3{\left({\cos }^{3}5x\right)}^{/}$ 

Tính ${\left({\sin }^{2}4x\right)}^{/}$: Áp dụng ${\left(u^{\alpha }\right)}^{/}$, với $u=\sin 4x,$ ta được:

$\begin{array}{l}{\left({\sin }^{2}4x\right)}^{/}=2\sin 4x.{\left(\sin 4x\right)}^{/}\\ =2\sin 4x.\cos 4x{\left(4x\right)}^{/}\\ =8.\sin 4x.c\text{os4x }=4\sin 8x.\end{array}$ 

Tương tự:  

$\begin{array}{l}{\left({\cos }^{3}5x\right)}^{/}=3{\cos }^{2}5x.{\left(\cos 5x\right)}^{/}\\ =3{\cos }^{2}5x.\left(-\sin 5x\right).{\left(5x\right)}^{/}\end{array}$

 $\begin{array}{l}=-15{\cos }^{2}5x.\sin 5x\\ =\frac{-15}{2}cos5x.\sin 10x.\end{array}$ 

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

Chọn D

Áp dụng ${\left(u^{\alpha }\right)}^{/}$, với $u=2+{\sin }^{2}2x.$ 

$\begin{array}{l}y\text{'}=3{\left(2+{\sin }^{2}2x\right)}^2{\left(2+{\sin }^{2}2x\right)}^{/}\\ =3{\left(2+{\sin }^{2}2x\right)}^2{\left({\sin }^{2}2x\right)}^{/}.\end{array}$ 

Tính ${\left({\sin }^{2}2x\right)}^{/},$ áp dụng ${\left(u^{\alpha }\right)}^{/},$ với $u=\sin 2x.$ 

$\begin{array}{l}{\left({\sin }^{2}2x\right)}^{/}=2.\sin 2x{\left(\sin 2x\right)}^{/}\\ =2.\sin 2x.\cos 2x{\left(2x\right)}^{/}\\ =2\sin 4x.\end{array}$ 

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

Chọn C

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left[2+\tan \left(x+\frac{1}{x}\right)\right]}^{\text{'}}}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\\ =\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\cdot {\left(x+\frac{1}{x}\right)}^{\text{'}}\end{array}$

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

Chọn B

$\begin{array}{l}{y}^{\text{'}}=2\cot \left(\cos x\right).{\left(\cot \left(\cos x\right)\right)}^{\text{'}}+\frac{{\left(\sin x\text{-}\frac{\pi }{2}\right)}^{\text{'}}}{2\sqrt{s\text{in}x-\frac{\pi }{2}}}\\ =2\cot \left(\cos x\right).\frac{-1}{{\sin }^2\left(\cos x\right)}.(\text{}c\text{osx)' +\hspace{0.17em}}\frac{\text{cosx}}{2\sqrt{s\text{in}x-\frac{\pi }{2}}}\\ =2\cot \left(\cos x\right).\frac{1}{{\sin }^2\left(\cos x\right)}.\sin x+\frac{\cos x}{2\sqrt{\sin x-\frac{\pi }{2}}}\end{array}$

Chọn  A

$\begin{array}{l}{y}^{\text{'}}={\left(\frac{\cos x}{2{\sin }^{2}x}\right)}^{\text{'}}\\ =\frac{2{\sin }^{2}x{\left(\cos x\right)}^{\text{'}}-\left(2{\sin }^{2}x\right)\text{'}.\cos x}{4{\sin }^{4}x}\\ =\frac{-2{\sin }^{3}x-2.2\sin x.\cos x.\cos x}{4{\sin }^{4}x}\end{array}$

$\begin{array}{l}=-\frac{{\sin }^{2}x+2{\text{cos}}^{2}x}{2{\sin }^{3}x}\\ =-\frac{({\sin }^{2}x+\text{}c{\text{os}}^{2}x)+{\text{cos}}^{2}x}{2{\sin }^{3}x}\\ =-\frac{1+{\text{cos}}^{2}x}{2{\sin }^{3}x}\end{array}$

Chọn D.

Bước đầu tiên ta áp dụng công thức ${\left(u^{\alpha }\right)}^{/}$ với $u=\frac{\sin x}{1+\cos x}$ 

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

Tính :

$\begin{array}{l}{\left(\frac{\sin x}{1+\cos x}\right)}^{/}\\ =\frac{{\left(\sin x\right)}^{/}\left(1+\cos x\right)-{\left(1+\cos x\right)}^{/}.\sin x}{{\left(1+\cos x\right)}^2}\\ =\frac{\cos x\left(1+\cos x\right)+{\sin }^{2}x}{{\left(1+\cos x\right)}^2}\end{array}$ 

 $\begin{array}{l}=\frac{\cos x+{\cos }^{2}x+{\sin }^{2}x}{{\left(1+\cos x\right)}^2}\\ =\frac{\text{cosx + 1}}{{(1+c\text{osx)}}^2}=\frac{1}{1+\cos x}\end{array}$.

Vậy

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

Chọn A

Áp dụng ${\left(\sin u\right)}^{/},$ với $u={\cos }^{2}x{\tan }^{2}x$ 

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

Tính ${\left({\cos }^{2}x.{\tan }^{2}x\right)}^{/},$ bước đầu sử dụng ${\left(u.v\right)}^{/},$ sau đó sử dụng ${\left(u^{\alpha }\right)}^{/}.$ 

$\begin{array}{l}{\left({\cos }^{2}x.{\tan }^{2}x\right)}^{/}\\ ={\left({\cos }^{2}x\right)}^{/}.{\tan }^{2}x+{\left({\tan }^{2}x\right)}^{/}.{\cos }^{2}x\end{array}$

$=2\cos x{\left(\cos x\right)}^{/}{\tan }^{2}x+2\tan x{\left(\tan x\right)}^{/}{\cos }^{2}x$

$\begin{array}{l}=-2\sin x\cos x{\tan }^{2}x+2\tan x\frac{1}{{\cos }^{2}x}{\cos }^{2}x\\ =-\sin 2x{\tan }^{2}x+2\tan x.\end{array}$ 

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

Chọn D

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

$\begin{array}{l}y\text{'}=\frac{\left(2\cos 2x-2\sin 2x\right)\left(2\sin 2x-\cos 2x\right)-\left(4\cos 2x+2\sin 2x\right)\left(\sin 2x+\cos 2x\right)}{{\left(2\sin 2x-\cos 2x\right)}^2}\\ =\frac{4.c{\text{os2x. sin2x - 2cos}}^{2}2x-4{\sin }^{2}2x+\text{}2.\sin 2x.c\text{os2x }}{{(2\sin 2x-\cos 2x)}^2}\\ -\frac{{\text{ ( 4cos2x . sin2x + 4cos}}^{2}2x+2{\sin }^{2}2x+2\sin 2x.c\text{os2x}}{{(\text{}2\sin 2x-c\text{os2x)}}^2}\end{array}$

$\begin{array}{l}y\text{'}=\frac{-6{\cos }^{2}2x-6{\sin }^{2}2x}{{\left(2\sin 2x-\cos 2x\right)}^2}\\ =\frac{-6}{{\left(2\sin 2x-\cos 2x\right)}^2}\end{array}$

Chọn C

Đầu tiên áp dụng ${\left(u^{\alpha }\right)}^{/},$ với $u=\sin \left(\cos \left({\tan }^{4}3x\right)\right)$ 

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

Sau đó áp dụng ${\left(\sin u\right)}^{/},$ với $u=\cos \left({\tan }^{4}3x\right)$ 

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

Áp dụng ${\left(\cos u\right)}^{/},$ với $u={\tan }^{4}3x.$ 

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

Áp dụng ${\left(u^{\alpha }\right)}^{/},$ với $u=\tan 3x$ 

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

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

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