28 câu Trắc nghiệm Các quy tắc tính đạo hàm có đáp án (phần 2)

Chọn C

Ta có :  .f ' (x)=4x  $\Rightarrow$f ' (-1)=-4

Chọn D

Ta có:

 $$\begin{gathered} f \text{'} ( x) = -4x^3+4.3x^2-3.2x+2.1+0 \\ = -4x^3+12x^2-6x+2 \end{gathered}$$ .

Nên

 $$\begin{gathered} f \text{'} (-1) =-4(-1)+12.1-6(-1)+2 \\ =24 \end{gathered}$$ .

Chọn C.

Ta có:$\begin{array}{l}{y}^{\text{'}}=4{\left(x^2+1\right)}^3\cdot {\left(x^2+1\right)}^{\text{'}}\\ =4{\left(x^2+1\right)}^3\cdot 2x=8x{\left(x^2+1\right)}^3\end{array}$

Nên y’(-1) = 8.(-1).( 1+ 1)³ = - 64

Chọn B.

Ta có $y\text{'}=3x^2-6x-9$

$$\begin{gathered} y\text{'}=0 \\ \iff 3x^2-6x-9=0 \\ \iff x=-1;x=3 \end{gathered}$$

Chọn D

ta có

 $$\begin{gathered} f\left(x\right) =\frac{x^2-2x+5}{x-1} \\ =x-1+\frac{4}{x-1} \\ \Rightarrow f\text{'}\left(x\right)=1-\frac{4}{{\left(x-1\right)}^2}\Rightarrow f\text{'}\left(-1\right)=0 \end{gathered}$$

Chọn D

Ta có : $f^{\text{'}}\left(x\right)=\frac{1}{2\sqrt{x^2}}.\left(x^2\right)\text{'}=\frac{x}{\sqrt{x^2}}$

Vì f '(x) không xác định tại x =  0

Do đó,   hàm số  không có đạo hàm tại x= 0.

Chọn B

Ta có 

$$\begin{gathered} y=-4x^3+4x \\ \Rightarrow y\text{'}=-12x^2+4 \end{gathered}$$

nên

$\begin{array}{l}y\text{'}\ge 0\iff -12x^2+\text{\hspace{0.17em}\hspace{0.17em}}4\ge 0\\ \iff x^2\le \frac{1}{3}\iff x\in \left[\frac{-1}{\sqrt{3}};\frac{1}{\sqrt{3}}\right]\end{array}$

Chọn C.

Ta có: $y\text{'}=m{x}^2-2mx+3m-1$

Nên $y\text{'}\le 0\forall x\iff m{x}^2-2mx+\text{}3m-1\le 0\forall x\left(1\right)$

m = 0 thì (1) trở thành: $-1\le 0$ đúng với $\forall x\in R$

$m\ne 0$, khi đó (1) đúng với $\forall x\in R\iff \left\{\begin{array}{l}a=m<0\\ ∆\text{'}\le 0\end{array}\right.$

$$\begin{gathered} \iff \left\{\begin{array}{l}m<0\\ m^2-m.(3m-1)=m(1-2m)\le 0\end{array}\right. \\ \iff \left\{\begin{array}{l}m<0\\ 1-2m\ge 0\end{array}\right. \\ \iff \left\{\begin{array}{l}m<0\\ m\le \frac{1}{2}\end{array}\right.\iff m<0 \end{gathered}$$

Vậy $m\le 0$ là những giá trị cần tìm.

Chọn D

Chọn B.

Đặt $u=x^3-2x^2$ thì $y=u^{2016}$, ${y^{\text{'}}}_u=2016.u^{2015},$${u^{\text{'}}}_x=3x^2-4x.$

Theo công thức tính đạo hàm của hàm số hợp, ta có: ${y^{\text{'}}}_x={y^{\text{'}}}_u.{u^{\text{'}}}_x$.

Vậy: y'=$2016.{{(x^3-2x^2)}^2}^{015}.(3x^2-4x).$ 

Chọn C

Áp dụng công thức đạo hàm 1 thương,ta có 

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

Chọn C

chọn C

Chọn A

Đáp án C

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

Đáp án B

Ta có:

$f\text{'}\left(x\right)={\left(2x^3+x-\sqrt{2}\right)}^{/}=6x^2+1$

$g\text{'}\left(x\right)={\left(3x^2+x+\sqrt{2}\right)}^{/}=6x+1$

$\begin{array}{l}f\text{'}\left(x\right)>g\text{'}\left(x\right)\iff 6x^2+1>6x+1\\ \iff 6x^2-6x>0\iff x\in \left(-\infty ;0\right)\cup \left(1;+\infty \right)\end{array}$

$\begin{array}{l}y\text{'}=\frac{(\text{}4x+1)\text{'}.\sqrt{x^2+2}-(4x+1).\left(\sqrt{x^2+\text{}2}\right)\text{'}}{x^2+2}\\ =\frac{4.\sqrt{x^2+2}-(4x+1)\text{}.\frac{x}{\sqrt{x^2+2}}}{x^2+\text{}2}=\frac{4(x^2+\text{}2)-(\text{}4x+1).x}{(x^2+\text{}2).\sqrt{x^2+\text{}2}}\\ =\frac{-x+\text{}8}{(x^2+\text{}2).\sqrt{x^2+\text{}2}}\end{array}$

ĐÁP ÁN D

Chọn B.

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

Để 

$$\begin{gathered} f\text{'}\left(x\right)=-2\iff x^2+x-2=-2 \\ \iff x^2+x=0\iff \left[\begin{array}{c}x=0\\ x=-1\end{array}\right. \end{gathered}$$

Chọn A

Bước đầu tiên sử dụng ${\left(u^{\alpha }\right)}^{/}$, với $u=\frac{2x+\text{}1}{x-1}$

 $\begin{array}{l}y\text{'}=3.{\left(\frac{2x+1}{x-1}\right)}^2.{\left(\frac{2x+1}{x-1}\right)}^{/}\\ =3.{\left(\frac{2x+1}{x-1}\right)}^2.\frac{2(x-1)-1.(2x+1)}{{(x-1)}^2}\\ =3.{\left(\frac{2x+1}{x-1}\right)}^2.\frac{-3}{{\left(x-1\right)}^2}\\ =-\frac{9{\left(2x+1\right)}^2}{{\left(x-1\right)}^4}.\end{array}$

Chọn  C

Đầu tiên sử dụng công thức ${\left(\frac{1}{u}\right)}^{/}$ với $u={\left(x^2-x+1\right)}^5$ 

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

Chọn B.

Sử dụng công thức ${\left(\sqrt{u}\right)}^{/}$  với  $u=1+2x-x^2$

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

Chọn A

Sử dụng công thức ${\left(\sqrt{u}\right)}^{/}$ với $u=\frac{x^2+1}{x}$ 

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

Chọn C

Sử dụng $\left(\frac{u}{v}\right)\text{'}=\frac{u\text{'}.v-u.v\text{'}}{v^2}$ ta được

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

Chọn A.

Đầu tiên áp dụng $\left(\sqrt{u}\right)\text{'}$  với $u={(x-2)}^3$ 

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

Chọn C

Ta có

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

$\begin{array}{l}f\text{'}\left(x\right)>g\text{'}\left(x\right)\\ \iff 6x^2-2x>3x^2+x\\ \iff 3x^2-3x>0\\ \iff x\in \left(-\infty ;0\right)\cup \left(1;+\infty \right)\end{array}$

Chọn B.

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

$y\text{'}\left(1\right)=\frac{1^2-4.1-2}{{(1-2)}^2}=-5$

Chọn A

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

$\Rightarrow y\text{'}\left(0\right)=\frac{2+\text{}0}{4-0}=\frac{1}{2}$

Chọn C

Ta có $y\text{'}=4-\frac{1}{2\sqrt{x}}$

$y\text{'}\left(x\text{}\right)\text{\hspace{0.17em}}=0\iff 4-\frac{1}{2\sqrt{x}}=0$

$\begin{array}{l}\iff 4=\frac{1}{2\sqrt{x}}\iff 8\sqrt{x}=1\\ \iff \sqrt{x}=\frac{1}{8}\iff x=\frac{1}{64}\end{array}$