10 câu Trắc nghiệm Đạo hàm cấp hai có đáp án có đáp án

Đáp án D

Ta có $y\text{'}=\frac{x\text{'}.(x-2)-x.(x-2)\text{'}}{{(x-2)}^2}=\frac{1.(x-2)-x.1}{{(x-2)}^2}=\frac{-2}{{(x-2)}^2}$ ;

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

Đáp án C

Ta có $y=x^6+\text{}3x^4+\text{}3x^2+\text{}1$

$\begin{array}{l}y\text{'}=6x^5+\text{}12x^3+6x\\ y"=30x^4+\text{}36x^2+\text{}6\\ y"\text{'}=120x^3+\text{}72x=24(5x^3+\text{}3x)\end{array}$

Đáp án C

Ta có $y\text{'}={\left(\sqrt{2x+5}\right)}^{\text{'}}=\frac{2}{2\sqrt{2x+5}}=\frac{1}{\sqrt{2x+5}}$ 

$y\text{'}\text{'}=-\frac{{\left(\sqrt{2x+5}\right)}^{\text{'}}}{2x+5}=-\frac{{\displaystyle \frac{2}{2\sqrt{2x+5}}}}{2x+5}=-\frac{1}{\left(2x+5\right)\sqrt{2x+5}}$.

Đáp án A

Ta có:

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

$\Rightarrow y"\left(1\right)=\frac{2}{{(1-3)}^3}=-\frac{1}{4}$.

Đáp án A

Ta có : $y\text{'}=2\cos 2x;y\text{'}\text{'}=-4\sin 2x$

 $y\text{'}\text{'}\text{'}=-8\cos 2x,y^{\left(4\right)}=16\sin 2x$

Suy ra $y\text{'}\text{'}\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=-8\cos \frac{2\mathrm{\pi }}{3}=4;y^{\left(4\right)}\left(\frac{\mathrm{\pi }}{4}\right)=16\sin \frac{\mathrm{\pi }}{2}=16$.

Đáp án D

Ta có gia tốc tức thời của chuyển động tại thời điểm t bằng đạo hàm cấp hai của phương trình chuyển động tại thời điểm t.

$s\text{'}={\left(t^3-3t^2+5t+2\right)}^{\text{'}}=3t^2-6t+5$

$s\text{'}\text{'}=6t-6\Rightarrow s^{\text{'}\text{'}}\left(3\right)=12$

Đáp án C

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

$\begin{array}{l}y"=\frac{4x.\sqrt{x{}^2+\text{}1}-(2x^2+1).\frac{x}{\sqrt{x^2+1}}}{x^2+1}\\ =\frac{4x(x^2+1)-(2x^2+1).x}{(x^2+1).\sqrt{x^2+1}}=\frac{2x^3+3x}{(x^2+1).\sqrt{x^2+1}}\end{array}$

Đáp án B

Ta có: $y\text{'}=5.{(2x+5)}^4.2=10{(2x+5)}^4$ ;

$y\text{'}\text{'}=10.4{(2x+5)}^3.{\left(2x+5\right)}^{\text{'}}=80{(2x+5)}^3$;

$y\text{'}\text{'}\text{'}=80.3{(2x+5)}^2.(2x+5)\text{'}=480{\left(2x+5\right)}^2$.

Đáp án D

Ta có: $y\text{'}=\frac{1}{{\cos }^{2}x}$ ;

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

Đáp án B

$\begin{array}{l}y=\frac{-2x^3+\text{}3x}{1-x}=2x^2+2x-1+\text{\hspace{0.17em}}\frac{1}{1-x}\\ y\text{'}=4x+\text{}2-\frac{1}{{(1-x)}^2}.(1-x)\text{'}=4x+2+\text{\hspace{0.17em}}\frac{1}{{(1-x)}^2}\\ y"=4-\frac{1}{{(1-x)}^4}\left[{(1- x)}^2\right]\text{'}=4-\frac{1}{{(1-x)}^4}.2(1-x).(-1)\\ =4+\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{{(1-x)}^3}\end{array}$