Hướng dẫn giải:

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

Hướng dẫn giải:

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

Hướng dẫn giải:

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

Hướng dẫn giải:

$f\text{'}\left(x\right)=2\cos \left(\frac{5\pi }{6}+x\right)$

$f\text{'}\left(\frac{\pi }{6}\right)=2\cos \pi =-2$

Hướng dẫn giải:

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

$f\text{'}\left(0\right)=\frac{1}{{\text{cos}}^2\left(\frac{-2\pi }{3}\right)}=\frac{1}{{\left(\frac{-1}{2}\right)}^2}=4$

Hướng dẫn giải:

Ta có: 

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

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

Hướng dẫn giải:

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

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

Hướng dẫn giải:

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

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

Hướng dẫn giải:

Ta có:

Hướng dẫn giải:

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

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

Hướng dẫn giải:

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

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

Hướng dẫn giải:

$\begin{array}{l}y=\frac{-c\text{osx}}{{\sin }^{3}x}+\frac{4}{3}\cot x\\ =-\cot x.\frac{1}{{\sin }^{2}x}+\frac{4}{3}\cot x\\ =-\cot x.(1+\text{}c{\text{ot}}^{2}x)+\text{\hspace{0.17em}}\frac{4}{3}\cot x\\ =-\cot x-{\cot }^{3}x+\text{\hspace{0.17em}}\frac{4}{3}\cot x\\ =-{\cot }^{3}x+\frac{1}{3}\cot x\end{array}$

Suy ra

$\begin{array}{l}f\text{'}\left(\frac{\pi }{3}\right)=\frac{3.{\left(\frac{1}{\sqrt{3}}\right)}^2}{{\left(\frac{\sqrt{3}}{2}\right)}^2}-\frac{1}{3.{\left(\frac{\sqrt{3}}{2}\right)}^2}\\ =\frac{4}{3}-\frac{4}{9}=\frac{8}{9}\end{array}$

Hướng dẫn giải:

$f^{\text{'}}\left(x\right)=\frac{-2\cos x\sin x\left(1+{\sin }^{2}x\right)-2\cos x\sin x{\cos }^{2}x}{{\left(1+{\sin }^{2}x\right)}^2}$
   $\begin{array}{l}=\frac{-2\cos x\sin x\left(1+{\sin }^{2}x+{\cos }^{2}x\right)}{{\left(1+{\sin }^{2}x\right)}^2}\\ =\frac{-4\cos x\sin x}{{\left(1+{\sin }^{2}x\right)}^2}\end{array}$
 $\Rightarrow f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{-8}{9}$

$f\left(\frac{\pi }{4}\right)-3f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{1}{3}+\frac{8}{3}=3$

Đáp án:
$\begin{array}{l}\left(ta{n}^2\frac{x}{2}\right)\text{'}=2tan\frac{x}{2}\left(tan\frac{x}{2}\right)\text{'}\\ =2tan\frac{x}{2}.\frac{\left(\frac{x}{2}\right)\text{'}}{co{s}^2\frac{x}{2}}\\ =2tan\frac{x}{2}.\frac{1}{2co{s}^2\frac{x}{2}}\end{array}$

$=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}.\frac{1}{co{s}^2\frac{x}{2}}=\frac{sin\frac{x}{2}}{co{s}^3\frac{x}{2}}$

Đáp án:

 Ta có:

$f\left(x\right)=\tan \left(x-\frac{2\pi }{3}\right)$

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

Đáp án:

$y=x(2x-1)(3x+2)(sinx-cosx)\text{'}$

$=(6x^3+x^2-2x)(sinx+cosx)$

$\Rightarrow y\text{'}=(6x^3+x^2-2x)\text{'}(sinx+cosx)+(6x^3+x^2-2x)(sinx+cosx)\text{'}$

$\begin{array}{l}y\text{'}=(18x^2+2x-2)(sinx+cosx)+(6x^3+x^2-2x)(cosx-sinx)\\ y\text{'}=sinx(18x^2+2x-2-6x^3-x^2+2x)+cosx(18x^2+2x-2+6x^3+x^2-2x)\\ y\text{'}=sinx(-6x^3+17x^2+4x-2)+cosx(6x^3+19x^2-2)\end{array}$

Hướng dẫn giải:

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 = sin2x

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

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

Hướng dẫn giải:

$\begin{array}{l}y={\left[\left({\cos }^{2}x-{\sin }^{2}x\right)\left({\cos }^{2}x+{\sin }^{2}x\right)\right]}^5\\ ={\left[\text{cos2x }.1\right]}^5={\left(\cos 2x\right)}^5.\end{array}$

Áp dụng  , với u= cos2x

$\begin{array}{l}y\text{'}=5.{\cos }^{4}2x.{\left(\cos 2x\right)}^{/}\\ =5.{\cos }^{4}2x.\left(-\sin 2x\right).{\left(2x\right)}^{/}\\ =-10{\cos }^{4}2x.\sin 2x.\end{array}$

Hướng dẫn giải:

$\begin{array}{l}y\text{'}=\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}\\ =\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x.{\cos }^{2}x}\\ =\frac{1.4}{{(2\sin x.\cos x)}^2}\\ =\frac{4}{{\sin }^{2}2x}\end{array}$

Hướng dẫn giải:

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

Hướng dẫn giải:

Sử dụng công thức đạo hàm hợp: $\left(u^n\right)\text{'}=n.u^{n-1}.u\text{'}$ và đạo hàm của hàm số lượng giác.

Ta có: 

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

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

$=\left(1+\tan x\right)\left(1+{\tan }^{2}x\right)$

Hướng dẫn giải: 

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

Hướng dẫn giải:

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

Hướng dẫn giải:

Ta có:  

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

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

Hướng dẫn giải: 

Ta có:  

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

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

Hướng dẫn giải:

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

Hướng dẫn giải:

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

$\begin{array}{l}=\sin x\left(2{\cos }^{2}x-{\sin }^{2}x\right)\\ =\sin x\left(3{\cos }^{2}x-1\right)\end{array}$

Hướng dẫn giải:

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

Hướng dẫn giải:

Áp dụng ${\left(u^{\alpha }\right)}^{/},$  với $u=\cos \left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)$  
$\begin{array}{l}y\text{'}=2.\cos \left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right).{\left[\cos \left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\right]}^{/}\\ =-2.\cos \left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right).\sin \left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right).{\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)}^{/}\end{array}$  
$y\text{'}=-\sin \left(2\frac{\sqrt{x}+1}{\sqrt{x}-1}\right).{\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)}^{/}.$

Tính   

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

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

Hướng dẫn giải: