Đáp án:

Ta có:

$\begin{array}{l}y\text{'}=\frac{3.(1-x)-(-1).(3x-2)}{{(1-x)}^2}\\ =\frac{1}{{(1-x)}^2}\\ \Rightarrow y\text{'}\text{'}=-\frac{{\text{[(1-x)}}^2\text{]}\text{'}}{{(1-x)}^4}\\ =-\frac{-2(1-x)}{{(1-x)}^4}\\ =\frac{2}{{(1-x)}^3}\end{array}$

Bất phương trình $y^{\text{'}\text{'}}>0$

$\begin{array}{l}\iff \frac{2}{{\left(1-x\right)}^3}>0\iff {\left(1-x\right)}^3>0\\ \iff 1-x>0\iff x<1\end{array}$

Đáp án:

Ta có 

Đáp án:

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

Đáp án:

$\begin{array}{l}y\text{'}=\frac{1}{co{s}^{2}x}\\ y\text{'}\text{'}=\frac{-\left(co{s}^{2}x\right)\text{'}}{co{s}^{4}x}\\ =\frac{-2cosx\left(cosx\right)\text{'}}{co{s}^{4}x}\\ =\frac{2sinx}{co{s}^{3}x}\end{array}$

Đáp án:

Ta có 

$y=x+\frac{1}{x+1}\Rightarrow y\text{'}=1-\frac{1}{{(x+1)}^2}$

$\begin{array}{l}\Rightarrow y\text{'}\text{'}=0-\frac{-2(x+\text{}1)}{{(x+\text{}1)}^4}=\frac{2}{{(x+1)}^3}\\ \Rightarrow y^{\left(3\right)}=2.\frac{-3{(x+\text{}1)}^2}{{(x+\text{}1)}^6}=\frac{-6}{{(x+1)}^4}\\ \Rightarrow y^{\left(4\right)}=-6.\frac{-4{(x\text{}+1)}^3}{{(x+\text{}1)}^8}=\frac{24}{{(x+1)}^5}\\ \Rightarrow y^{\left(5\right)}=24.\frac{-5.{(x+1)}^4}{{(x+1)}^{10}}=\frac{-120}{{(x+1)}^6}\end{array}$

Đáp án:

Cách 1:

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

$\begin{array}{l}y\text{'}\text{'}=6{\left(x^2+1\right)}^2+6x.2(x^2+1).2x\\ =6{\left(x^2+1\right)}^2+24x^2(x^2+1)\\ y\text{'}\text{'}\text{'}=12(x^2+1).2x+24.2x.(x^2+1)+24x^2.2x\\ =24x(x^2+1)+48x(x^2+1)+48x^3\\ =24x.\text{[}{x}^2+1+2(x^2+1)+2x^2\text{]}\\ =24x.(5x^2+3)\end{array}$

Cách 2:

$\begin{array}{l}y={\left(x^2+1\right)}^3\\ =x^6+3x^4+3x^2+1\\ y\text{'}=6x^5+12x^3+6x\end{array}$

$\begin{array}{l}y\text{'}\text{'}=30x^4+36x^2+6\\ y\text{'}\text{'}\text{'}=120x^3+72x\\ =24x(5x^2+3)\end{array}$

Đáp án:

$\begin{array}{l}f\text{'}\left(x\right)=5{\left(2x+5\right)}^{4}2x+5)\text{'}\\ =10{\left(2x+5\right)}^4\\ f\text{'}\text{'}\left(x\right)=40{\left(2x+5\right)}^3(2x+5)\text{'}\\ =80{\left(2x+5\right)}^3\\ f\text{'}\text{'}\text{'}\left(x\right)=240{\left(2x+5\right)}^2(2x+5)\text{'}\\ =480{\left(2x+5\right)}^2\end{array}$

Đáp án:

Ta có 

$\begin{array}{l}{y}^{\text{'}}=-2\sin x\cos x\\ =-\sin 2x\\ \Rightarrow y\text{'}\text{'}=-2cos2x\\ \Rightarrow y\text{'}\text{'}\text{'}=4sin2x\end{array}$

$\Rightarrow y\text{'}\text{'}\text{'}\left(\frac{\pi }{3}\right)=2\sqrt{3}.$

Đáp án:

Hàm số viết lại: $y=x^4-2x^2+1$

Ta có

$\begin{array}{l}y\text{'}=4x^3-4x\\ y\text{'}\text{'}=12x^2-4\\ y\text{'}\text{'}\text{'}=24x\\ y^{\left(4\right)}=24\end{array}$

Khi đó

$\begin{array}{l}M=y^{\left(4\right)}+2xy\text{'}\text{'}\text{'}-4y\text{'}\text{'}\\ =24+2x.24x-4(12x^2-4)=40.\end{array}$

Đáp án:

$\begin{array}{l}y\text{'}=\frac{(2x+5)\text{'}}{2\sqrt{2x+5}}\\ =\frac{1}{\sqrt{2x+5}}={\left(2x+5\right)}^{\frac{-1}{2}}\\ y\text{'}\text{'}=-\frac{1}{2}.{\left(2x+5\right)}^{\frac{-1}{2}-1}(2x+5)\text{'}\\ =-\frac{1}{2}{\left(2x+5\right)}^{\frac{-3}{2}}.2\\ =-\frac{1}{{(2x+5)}^{\frac{3}{2}}}\\ =-\frac{1}{(2x+5)\sqrt{2x+5}}\end{array}$

Đáp án:

$\begin{array}{l}y\text{'}=\sqrt{x^2+1}+x.\frac{2x}{2\sqrt{x^2+1}}\\ =\frac{x^2+1+x^2}{\sqrt{x^2+1}}=\frac{2x^2+1}{\sqrt{x^2+1}}\end{array}$

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

$\begin{array}{l}=\frac{4x^3+4x-2x^3-x}{\left(x^2+1\right)\sqrt{x^2+1}}\\ =\frac{2x^3+3x}{(x^2+1)\sqrt{x^2+1}}\end{array}$

Đáp án:

Ta có

$\begin{array}{l}y\text{'}=2cos2x+2sin2x\\ \Rightarrow y\text{'}\text{'}=-4cos2x+4sin2x\end{array}$  

Phương trình

$\begin{array}{l}y\text{'}\text{'}=0\iff -4cos2x+4sin2x=0\\ \iff sin2x-cos2x=0\end{array}$

$\begin{array}{l}\iff \sqrt{2}sin(2x-\frac{\pi }{4})=0\\ \iff sin(2x-\frac{\pi }{4})=0\\ \iff 2x-\frac{\pi }{4}=k\pi \\ \iff x=\frac{\pi }{8}+\frac{k\pi }{2},k\in Z\end{array}$

Đáp án:

Ta có 

$\begin{array}{l}y=\frac{1}{2}{x}^2+x+1\\ \Rightarrow y^{\text{'}}=x+1\end{array}$

và $y\text{'}\text{'}=1$

Khi đó 

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

$=x^2+2x+1-x^2-2x-2=-1$

Đáp án:

$h\text{'}\left(x\right)=15{(x+1)}^2+4$

$\begin{array}{l}h\text{'}\text{'}\left(x\right)=30(x+1)=0\\ \iff x=-1\end{array}$

Đáp án:

Ta có

$\begin{array}{l}f\text{'}\left(x\right)=\text{\hspace{0.17em}\hspace{0.17em}}\frac{2(x+\text{\hspace{0.33em}\hspace{0.33em}}1)-(2x-1).1}{{(x+1)}^2}\\ =\frac{3}{{(x+1)}^2}\end{array}$

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

Phương trình

$\begin{array}{l}f\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)\\ \iff \frac{3}{{(x+1)}^2}=\frac{-6}{{(x+1)}^3}\end{array}$

$\iff \left\{\begin{array}{l}\frac{-2}{x+1}=1\\ x\ne -1\end{array}\right.\iff x=-3.$

Đáp án:

Đáp án A:

$\begin{array}{l}y=\frac{1}{cosx}\\ y\text{'}=\frac{-\left(cosx\right)\text{'}}{co{s}^{2}x}=\frac{sinx}{co{s}^{2}x}\\ y\text{'}\text{'}=\frac{cosx.co{s}^{2}x-sinx.2cosx\left(cosx\right)\text{'}}{{\left(co{s}^{2}x\right)}^2}\\ =\frac{co{s}^{3}x+2si{n}^{2}xcosx}{co{s}^{4}x}=\frac{co{s}^{2}x+2si{n}^{2}x}{co{s}^{3}x}\end{array}$

Đáp án B:

$\begin{array}{l}y=-\frac{1}{cosx}\\ y\text{'}=\frac{\left(cosx\right)\text{'}}{co{s}^{2}x}=-\frac{sinx}{co{s}^{2}x}\\ y\text{'}\text{'}=-\frac{cosx.co{s}^{2}x-sinx.2cosx\left(cosx\right)\text{'}}{co{s}^{4}x}\\ =-\frac{co{s}^{3}x+2si{n}^{2}xcosx}{co{s}^{4}x}=-\frac{co{s}^{2}x-2si{n}^{2}x}{co{s}^{3}x}\end{array}$

Đáp án C:

$\begin{array}{l}y=cotx\\ y\text{'}=-\frac{1}{si{n}^{2}x}\end{array}$

$\begin{array}{l}\mathrm{y}\text{'}=\frac{2\mathrm{sinx}\left(\mathrm{sinx}\right)\text{'}}{{\sin }^4\mathrm{x}}\\ =\frac{2\mathrm{sinxcosx}}{{\sin }^4\mathrm{x}}=\frac{2\mathrm{cosx}}{{\sin }^3\mathrm{x}}\end{array}$

Đáp án D:

$\begin{array}{l}y=tanx\\ y\text{'}=\frac{1}{co{s}^{2}x}\end{array}$

$\begin{array}{l}y\text{'}\text{'}=\frac{-2cosx\left(cosx\right)\text{'}}{co{s}^{4}x}\\ =\frac{2sinxcosx}{co{s}^{4}x}=\frac{2sinx}{co{s}^{3}x}\end{array}$

Đáp án:

$\begin{array}{l}f\text{'}\left(x\right)=5.{\left(ax+b\right)}^4(ax+b)\text{'}=5a{\left(ax+b\right)}^4\\ f\text{'}\text{'}\left(x\right)=5a.4.{\left(ax+b\right)}^3(ax+b)\text{'}=20a^2{\left(ax+b\right)}^3\\ f\text{'}\text{'}\text{'}\left(x\right)=20a^2.3.{\left(ax+b\right)}^2(ax+\text{}b)\text{'}=60a^3{\left(ax+b\right)}^2\end{array}$

$\begin{array}{l}{f}^{\left(4\right)}\left(x\right)=120a^4(ax+b)\\ f^{\left(5\right)}\left(x\right)=120a^4.(ax+b)\text{'}=120a^5\end{array}$

$f^{\left(6\right)}\left(x\right)=0$

$\begin{array}{l}\Rightarrow f^{\left(10\right)}\left(x\right)=0\forall x\in R\\ \Rightarrow f^{\left(10\right)}\left(1\right)=0\end{array}$

Đáp án:

$y\text{'}=cosx=sin(x+\frac{\pi }{2})\Rightarrow$Đáp án A đúng.

$y\text{'}\text{'}=-sinx=sin(x+\pi )\Rightarrow$Đáp án B đúng.

$y\text{'}\text{'}\text{'}=-cosx=sin(x+\frac{3\pi }{2})\Rightarrow$Đáp án C đúng.

Đáp án

$\begin{array}{l}y\text{'}=2cos2x;\\ y\text{'}\text{'}=-4sin2x=-4y\\ \iff 4y+y\text{'}\text{'}=0\end{array}$

Đáp án:

Ta có :

$\begin{array}{l}a=v\text{'}=(s\text{'})\text{'}=s\text{'}\text{'}\\ s\text{'}=3t^2-4t+4\\ s\text{'}\text{'}=6t-4=a\\ a\left(2\right)=6.2-4=8(m/s^2)\end{array}$

Lời giải

$\begin{array}{l}y\text{'}=3x^2-12x+\text{\hspace{0.17em}}40\\ y"=6x-12\end{array}$

Đáp án:

Ta có 

$\begin{array}{l}v\left(t\right)=s\text{'}\left(t\right)=3t^2+8t\\ \Rightarrow a\left(t\right)=v\text{'}\left(t\right)=6t+8\end{array}$

Thời điểm vận tốc của vật bằng $11 m/s$

$\begin{array}{l}\Rightarrow v\left(t\right)=11\\ \iff 3t^2+8t=11\\ \iff \left[\begin{array}{l}t=1\\ t=-\frac{11}{3}\left(L\right)\end{array}\right.\end{array}$

Với $t=1\Rightarrow a\left(1\right)=6.1+8=14 m/s^2$

Đáp án:

Ta có 

$\begin{array}{l}v\left(t\right)=s\text{'}\left(t\right)=3t^2-6t-9\\ \Rightarrow a\left(t\right)=v\text{'}\left(t\right)=6t-6.\end{array}$

Thời điểm vận tốc bị triệt tiêu: 

$\begin{array}{l}v\left(t\right)=0\\ \iff 3t^2-6t-9=0\\ \iff \left[\begin{array}{l}t=-1\left(l\right)\\ t=3\end{array}\right.\end{array}$

Với $t=3\Rightarrow a\left(3\right)=6.3-6=12m/s^2$

Đáp án:

$\begin{array}{l}f\text{'}\left(x\right)=-2sin(2x-\frac{\pi }{3})\\ f\text{'}\text{'}\left(x\right)=-4cos(2x-\frac{\pi }{3})\\ f\text{'}\text{'}\text{'}\left(x\right)=8sin(2x-\frac{\pi }{3})\\ f^{\left(4\right)}\left(x\right)=16cos(2x-\frac{\pi }{3})\end{array}$

$f^{\left(4\right)}\left(x\right)=-8\iff cos(2x-\frac{\pi }{3})=-\frac{1}{2}$

$\begin{array}{l}\iff \left[\begin{array}{l}2x-\frac{\pi }{3}=\frac{2\pi }{3}+k2\pi \\ 2x-\frac{\pi }{3}=-\frac{2\pi }{3}+k2\pi \end{array}\right.\\ \iff \left[\begin{array}{l}x=\frac{\pi }{2}+k\pi \\ x=-\frac{\pi }{6}+k\pi \end{array}\right.(k\in Z)\\ x\in [0;\frac{\pi }{2}]\Rightarrow x=\frac{\pi }{2}\end{array}$

Lời giải

$\begin{array}{l}y\text{'}=10.{\left(2x+5\right)}^9.(2x+5)\text{'}=20.{\left(2x+5\right)}^9\\ y"=\mathrm{20.9.}{\left(2x+5\right)}^8.(2x+\text{}5)\text{'}=360.{\left(2x+5\right)}^8\end{array}$

Đáp án:

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

Đáp án:

$\begin{array}{l}y=\frac{x-3}{x+4}=\frac{x+4-7}{x+4}\\ =\frac{x+4}{x+4}-\frac{7}{x+4}\\ =1-\frac{7}{x+4}\end{array}$

$\begin{array}{l}y\text{'}=(1-\frac{7}{x+4})\text{'}\\ =-\frac{-7}{{(x+4)}^2}\\ =\frac{7}{{(x+4)}^2}\\ \Rightarrow y\text{'}\text{'}=7.\frac{-\text{[}{(x+\text{}4)}^2\text{]}\text{'}}{{(x+4)}^4}\\ =\frac{-7.2(x+\text{}4)}{{(x+\text{}4)}^4}=\frac{-14}{{(x+\text{}4)}^3}\end{array}$

Ta có:

$2{(y\text{'})}^2=2.{\left[\frac{7}{{(x+4)}^2}\right]}^2=\frac{98}{{(x+4)}^4}$

$\begin{array}{l}\Rightarrow 2{\left(y\text{'}\right)}^2:y\text{'}\text{'}\\ =\frac{98}{{(x+4)}^4}:[-\frac{14}{{(x+4)}^3}]\\ =\frac{98}{{(x+4)}^4}.\frac{{(x+4)}^3}{-14}\\ =\frac{-7}{x+4}=y-1\end{array}$

$\begin{array}{l}\Rightarrow 2{\left(y\text{'}\right)}^2:y\text{'}\text{'}=y-1\\ \Rightarrow 2{\left(y\text{'}\right)}^2=(y-1)y\text{'}\text{'}\end{array}$

Đáp án:

Ta có $f^{\text{'}}\left(x\right)=3x^2-4x+1$

và $f\text{'}\text{'}\left(x\right)=6x-4$

Khi đó $f\text{'}\left(\sqrt{2}\right)=7-4\sqrt{2}$

và $f^{\text{'}\text{'}}\left(\sqrt{2}\right)=6\sqrt{2}-4$.

Suy ra 

$M=7-4\sqrt{2}+\frac{2}{3}(6\sqrt{2}-4)=\frac{13}{3}$

Đáp án:

Ta có 

$f\text{'}\left(x\right)=sinx+xcosx$

$\begin{array}{l}\Rightarrow f\text{'}\text{'}\left(x\right)=cosx+cosx-xsinx=2cosx-f\left(x\right)\\ \Rightarrow f\text{'}\text{'}\text{'}\left(x\right)=-2sinx-f\text{'}\left(x\right)\end{array}$

Khi đó 

$f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)+f\text{'}\text{'}\text{'}\left(x\right)$

$\begin{array}{l}=f\left(x\right)+f\text{'}\left(x\right)+[2cosx-f(x\left)\right]\\ +[-2sinx-f\text{'}(x\left)\right]=2(cosx-sinx)\\ \Rightarrow P=f\left(\frac{\pi }{2}\right)+f\text{'}\left(\frac{\pi }{2}\right)+f\text{'}\text{'}\left(\frac{\pi }{2}\right)+f\text{'}\text{'}\text{'}\left(\frac{\pi }{2}\right)\\ =\text{}2\left(\text{cos}\frac{\pi }{2}-\sin \frac{\pi }{2}\right)=-2\end{array}$

Đáp án:

Ta có 

$f\text{'}\left(x\right)=4x-16sinx+2sin2x$

$\begin{array}{l}\Rightarrow f\text{'}\text{'}\left(x\right)=4-16cosx+4cos2x\\ \Rightarrow f^{\text{'}\text{'}}\left(\pi \right)=4-16.(-1)+\text{\hspace{0.17em}\hspace{0.17em}}4.1=24\end{array}$