23 câu Trắc nghiệm Đạo hàm của các hàm số lượng giác có đáp án (phần 2)

Chọn A

Ta áp dụng đạo hàm của 1 tích :

$\begin{array}{l}y\text{'}=\left(x\right)\text{'}.c\text{osx + x. (cosx)' }\\ \text{=1.cosx + x. (- sinx})\\ =c\text{osx- x.sin x}\end{array}$

Chọn D

Bước đầu tiên áp dung công thức ${\left(u^{\alpha }\right)}^{/}$với $u=\sin \left(2x+1\right)$ 

  Vậy

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

* Tính ${\left(\sin \left(2x+1\right)\right)}^{/}$: Áp dụng ${\left(\sin u\right)}^{/}$, với $u=\left(2x+1\right)$ 

Ta được: 

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

 $\begin{array}{l}\Rightarrow y\text{'}=3.{\sin }^2\left(2x+1\right).2\cos \left(2x+1\right)\\ =6{\sin }^2\left(2x+1\right)\cos \left(2x+1\right).\end{array}$

Chọn D.

Áp dụng công thức ${\left(\sin u\right)}^{/}$ Với $u=\sqrt{2+x^2}$

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

Chọn A.

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

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

Chọn C

$\begin{array}{l}f\text{'}\left(x\right)=\left[\frac{2}{\cos \left(\pi x\right)}\right]\text{'}\\ =2.\left(\cos \left(\pi x\right)\right)\text{'}.\frac{-1}{{\cos }^2\left(\pi x\right)}\\ =2\pi \left[-\sin \left(\pi x\right)\right].\frac{-1}{{\cos }^2\left(\pi x\right)}\\ =2.\pi \frac{\sin \left(\pi x\right)}{{\cos }^2\left(\pi x\right)}\end{array}$

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

Chọn B

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

$\begin{array}{l}y\text{'}\left(\frac{\pi }{3}\right)\\ =-3\sin 3\frac{\pi }{3}.\sin 2\frac{\pi }{3}+2\cos 3\frac{\pi }{3}.\cos 2\frac{\pi }{3}\\ =-\mathrm{3.0.}\frac{\sqrt{3}}{2}+\text{}2.(-1).\frac{-1}{2}=1\end{array}$

Chọn D

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

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

Chọn B

$\begin{array}{l}{f}^{\text{'}}\left(x\right)=\frac{1}{{\cos }^2\left(x-\frac{2\pi }{3}\right)}.{\left(x-\frac{2\pi }{3}\right)}^{\text{'}}\\ =\frac{1}{{\cos }^2\left(x-\frac{2\pi }{3}\right)}\\ \Rightarrow f^{\text{'}}\left(0\right)=\frac{1}{\frac{1}{4}}=4\end{array}$

Chọn  B

$\begin{array}{l}f\text{'}\left(x\right)=\frac{-\sin x.\left(1+2\sin x\right)-\cos x\mathrm{.2.}\cos x}{{\left(1+2\sin x\right)}^2}\\ =\frac{-\mathrm{s}{\text{inx - 2sin}}^{2}x-2c{\text{os}}^{2}x}{{(1+2\sin x)}^2}\\ =\frac{-\mathrm{s}\text{inx}-2({\sin }^{2}x+\text{}c{\text{os}}^{2}x)}{{(1\text{}+2\sin x)}^2}\\ =\frac{-\sin x-2}{{\left(1+2\sin x\right)}^2}\end{array}$

Chọn D

Áp dụng công thức đạo hàm của hàm hợp với $y=\frac{1}{u};u=c\text{os 3x}$ ta được 

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

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

Chọn D

Áp dụng công thức đạo hàm của hàm hợp :(sin u)’ = cosu.u’ với $u=\pi .\sin x$

 ta được 

$\begin{array}{l}{y}^{\text{'}}=(\pi .\sin x{)}^{\text{'}}.\cos (\pi .\sin x)\\ =\pi .\cos x.\cos (\pi .\sin x)\end{array}$

$\begin{array}{l}\Rightarrow y^{\text{'}}\left(\frac{\pi }{6}\right)\\ =\pi .\cos \frac{\pi }{6}.\cos \left(\pi .\sin \frac{\pi }{6}\right)\\ =\pi .\frac{\sqrt{3}}{2}.\cos \left(\pi .\frac{1}{2}\right)\\ =\frac{\sqrt{3}.\pi }{2}.\cos \frac{\pi }{2}=0\end{array}$

Chọn B

$\begin{array}{l}y\text{'}=\frac{1}{2\sqrt{\tan x\text{}+\text{\hspace{0.17em}}\cot x}}.(\text{}\tan x+\cot x)\text{'}\\ =\frac{1}{2\sqrt{\tan x\text{}+\text{\hspace{0.17em}}\cot x}}.\left(\frac{1}{{\text{cos}}^{2}x}-\frac{1}{{\sin }^{2}x}\right)\\ \Rightarrow y\text{'}\left(\frac{\pi }{4}\right)=\frac{1}{2\sqrt{1+1}}.\left(\frac{1}{\frac{1}{2}}-\frac{1}{\frac{1}{2}}\right)=0\end{array}$

Chọn A

$\begin{array}{l}{f}^{\text{'}}\left(x\right)=\frac{{\left(\cos x\right)}^{\text{'}}\left(1-\mathrm{s}\text{in}x\right)-(1-\mathrm{s}\text{in}x{)}^{\text{'}}cosx}{{\left(1-\mathrm{s}\text{in}x\right)}^2}\\ =\frac{-\sin x.(1-\mathrm{s}\text{inx})+c\text{osx.cosx}}{{(1-\mathrm{s}\text{inx})}^2}\\ =\frac{-\sin x+{\sin }^{2}x+c{\text{os}}^{2}x}{{(1-\mathrm{s}\text{inx})}^2}\\ =\frac{-\mathrm{s}\text{inx}+1}{{(1-\mathrm{s}\text{inx})}^2}=\frac{1}{1-\mathrm{s}\text{in}x}\\ \Rightarrow f^{\text{'}}\left(\frac{\pi }{6}\right)-f^{\text{'}}\left(-\frac{\pi }{6}\right)\\ =\frac{1}{1-\frac{1}{2}}-\frac{1}{1+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}}=\frac{4}{3}\end{array}$

Chọn B

$\begin{array}{l}y\text{'}=\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}\\ =\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x.{\cos }^{2}x}\\ =\frac{1}{{(\mathrm{s}\text{inx. cosx)}}^2}\\ =\frac{1}{{\left(\frac{\sin 2x}{2}\right)}^2}=\frac{4}{{\sin }^{2}2x}\end{array}$

Chọn D

Ta có:  

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

Theo giả thiết Ta có:  $y^{\text{'}}=0\iff \sin \left(\frac{2\pi }{3}+2x\right)=0$

$\begin{array}{l}\iff \frac{2\pi }{3}+2x=k\pi \\ \iff x=-\frac{\pi }{3}+\frac{k\pi }{2}\left(k\in \mathbb{Z}\right)\end{array}$

Chọn B

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

Chọn A

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

Chọn C

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

Suy ra $y^{\text{'}}=\cos x-\sin x+\cos 2x$

Chọn A

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

$\begin{array}{l}f\text{'}\left(\frac{{\pi }^2}{16}\right)=\frac{1}{2\sqrt{{\left(\frac{\pi }{4}\right)}^2}}\left(\cos \sqrt{{\left(\frac{\pi }{4}\right)}^2}-\sin \sqrt{{\left(\frac{\pi }{4}\right)}^2}\right)\\ =\frac{1}{2.\frac{\pi }{4}}\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)=0\end{array}$

chọn C

$\begin{array}{l}y=\sqrt{\tan x+\cot x}\\ \Rightarrow y^2=\tan x+\cot x\\ \Rightarrow y\text{'}.2y=\frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}\end{array}$

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

$\begin{array}{l}f\text{'}\left(\frac{\pi }{4}\right)\\ =\frac{1}{2\sqrt{\tan \frac{\pi }{4}+\cot \frac{\pi }{4}}}\left(\frac{1}{{\cos }^2\left(\frac{\pi }{4}\right)}-\frac{1}{{\sin }^2\left(\frac{\pi }{4}\right)}\right)\\ =\frac{1}{2\sqrt{2}}\left(2-2\right)=0\end{array}$

Chọn C

$\begin{array}{l}y=\frac{1}{\sqrt{\sin x}}\\ \Rightarrow y^2=\frac{1}{\sin x}\\ \Rightarrow y\text{'}2y=\frac{-\cos x}{{\sin }^{2}x}\end{array}$

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

$\begin{array}{l}f\text{'}\left(\frac{\pi }{2}\right)=\frac{-\sqrt{\sin \left(\frac{\pi }{2}\right)}}{2}.\frac{\cos \left(\frac{\pi }{2}\right)}{{\sin }^2\left(\frac{\pi }{2}\right)}\\ =\frac{-1}{2}.\frac{0}{1}=0\end{array}$

Chọn C

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

$y^{\text{'}}\left(\frac{\pi }{6}\right)=\frac{1}{1-\sin \frac{\pi }{6}}=2$

Chọn C

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

$\begin{array}{l}=\frac{-2\cos x\sin x\left(1+{\sin }^{2}x+{\cos }^{2}x\right)}{{\left(1+{\sin }^{2}x\right)}^2}\\ =\frac{-4\cos x\sin x}{{\left(1+{\sin }^{2}x\right)}^2}\end{array}$

$\Rightarrow f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{-8}{9}$