35 câu Dạng 2: Đạo hàm cấp cao

Ta có  $y^{\text{'}}=\frac{7}{{\left(x+2\right)}^2}\Rightarrow {y^{\text{'}}}^{\text{'}}=\frac{-7{\left[{\left(x+2\right)}^2\right]}^{\text{'}}}{{\left(x+2\right)}^4}=\frac{-14}{{\left(x+2\right)}^3}$

 $\Rightarrow {{y^{\text{'}}}^{\text{'}}}^{\text{'}}=\frac{14{\left[{\left(x+2\right)}^3\right]}^{\text{'}}}{{\left(x+2\right)}^6}=\frac{42}{{\left(x+2\right)}^4}\Rightarrow y^{\left(4\right)}=\frac{-42{\left[{\left(x+2\right)}^4\right]}^{\text{'}}}{{\left(x+2\right)}^8}=\frac{-168}{{\left(x+2\right)}^5}$.

Ta có  $y={\sin }^{2}2x=\frac{1}{2}\left(1-\cos 4x\right)$

$\Rightarrow y^{\text{'}}=2\sin 4x\Rightarrow {y^{\text{'}}}^{\text{'}}=8\cos 4x\Rightarrow {{y^{\text{'}}}^{\text{'}}}^{\text{'}}=-32\sin 4x$

 $\Rightarrow y^{\left(4\right)}=-128\cos 4x\Rightarrow y^{\left(5\right)}=512\sin 4x$

Ta có: $y^{\text{'}}=\frac{-7}{{\left(x-2\right)}^2},{y^{\text{'}}}^{\text{'}}=\frac{7.2}{{\left(x-2\right)}^3},{{y^{\text{'}}}^{\text{'}}}^{\text{'}}=\frac{-\mathrm{7.2.3}}{{\left(x-2\right)}^4}$ .

Bằng quy nạp ta chứng minh  $y^{\left(n\right)}=\frac{{\left(-1\right)}^n\mathrm{.7.}n!}{{\left(x-2\right)}^{n+1}}$ .  $\left(2\right)$

Với  $n=1$ ta thấy  $\left(2\right)$ đúng.

Giả sử  $\left(2\right)$ đúng với  $n=k$, tức là  $y^{\left(k\right)}=\frac{{\left(-1\right)}^k\mathrm{.7.}k!}{{\left(x-2\right)}^{k+1}}$ .

Ta có:   $y^{\left(k+1\right)}={\left(\frac{{\left(-1\right)}^k\mathrm{.7.}k}{{\left(x-2\right)}^{k+1}}\right)}^{\text{'}}=-\frac{{\left(-1\right)}^k\mathrm{.7.}k!.\left(k+1\right)}{{\left(x-2\right)}^{k+2}}=\frac{{\left(-1\right)}^{k+1}\mathrm{.7.}\left(k+1\right)!}{{\left(x-2\right)}^{k+2}}$.

Do đó  $\left(2\right)$ đúng với mọi số tự nhiên $n$.

Vậy theo nguyên lí quy nạp ta có công thức đạo hàm cấp cao của hàm số

 $y=\frac{3x+1}{x-2}$ là  $y^{\left(n\right)}=\frac{{\left(-1\right)}^n\mathrm{.7.}n!}{{\left(x-2\right)}^{n+1}}$ .

Ta có

  $y^{\text{'}}={\left(x\sin x\right)}^{\text{'}}\iff y\text{'}=x^{\text{'}}.\sin x+x.{\left(\sin x\right)}^{\text{'}}$

$\iff y^{\text{'}}=\sin x+x\cos x$

${y^{\text{'}}}^{\text{'}}=\left(\sin x+x\cos x\right)\text{'}={\left(\sin x\right)}^{\text{'}}+{\left(x\cos x\right)}^{\text{'}}$

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

Ta có  $x.{y^{\text{'}}}^{\text{'}}-2\left(y^{\text{'}}-\sin x\right)+xy=0$

$\iff x\left(2\cos x-x\sin x\right)-2\left(\sin x+x\cos x-\sin x\right)+x^2\sin x=0$

$\iff 2x\cos x-x^2\sin x-2x\cos x+x^2\sin x=0$

$\iff 0=0$

(điều phải chứng minh).

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

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

$=\frac{-\sqrt{2x-x^2}-\frac{1-x}{\sqrt{2x-x^2}}.\left(1-x\right)}{{\left(\sqrt{2x-x^2}\right)}^2}$

$=\frac{-\left(2x-x^2\right)-{\left(1-x\right)}^2}{\sqrt{2x-x^2}.{\left(\sqrt{2x-x^2}\right)}^2}=\frac{-1}{{\left(\sqrt{2x-x^2}\right)}^3}$.

Ta có $y^3.{y^{\text{'}}}^{\text{'}}+1=0\iff {\left(\sqrt{2x-x^2}\right)}^3.\frac{-1}{{\left(\sqrt{2x-x^2}\right)}^3}+1=0\iff -1+1=0$

(điều phải chứng minh).

Ta có:  $y=\frac{\left(\sin x+\cos x\right)\left({\sin }^{2}x+{\cos }^{2}x-\sin x\cos x\right)}{1-\sin x\cos x}$

$=\frac{\left(\sin x\cos x\right)\left(1-\sin x\cos x\right)}{1-\sin x\cos x}=\sin x+\cos x$

$\Rightarrow y^{\text{'}}=\cos x-\sin x\Rightarrow {y^{\text{'}}}^{\text{'}}=-\sin x-\cos x$.

Ta có  ${y^{\text{'}}}^{\text{'}}+y=0\iff -\sin x-\cos x+\sin x+\cos x=0\iff 0=0$ (điều phải chứng minh).

Đáp án C

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

Suy ra:  ${f^{\text{'}}}^{\text{'}}\left(x\right)=6x-2$. Suy ra  ${f^{\text{'}}}^{\text{'}}\left(1\right)=4$.

Đáp án A

Ta có  $f^{\text{'}}\left(x\right)=3\sin 3x$, suy ra  ${f^{\text{'}}}^{\text{'}}\left(x\right)=-9\sin 3x$.

Do đó  ${f^{\text{'}}}^{\text{'}}\left(-\frac{\pi }{2}\right)=-9\sin \left(-\frac{3\pi }{2}\right)=-9$.

Đáp án B

Ta có  $y^{\text{'}}=f^{\text{'}}\left(x\right)={\left(x\sin x-3\right)}^{\text{'}}=\sin x+x\cos x$

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

Đáp án D

Ta có

$y^{\text{'}}=\cos x=\sin \left(x+\frac{\pi }{2}\right)\Rightarrow {y^{\text{'}}}^{\text{'}}=\sin \left(x+\frac{\pi }{2}+\frac{\pi }{2}\right)=\sin \left(x+\pi \right)\Rightarrow {{y^{\text{'}}}^{\text{'}}}^{\text{'}}=\sin \left(x+\pi +\frac{\pi }{2}\right)=\sin \left(x+\frac{3\pi }{2}\right)$

$\Rightarrow y^{\left(4\right)}=\sin \left(x+\frac{3\pi }{2}+\frac{\pi }{2}\right)=\sin \left(x+2\pi \right)=\sin x$.

Ta có  $\sin \left(2\pi -x\right)=-\sin x\ne y^{\left(4\right)}$.

Đáp án D

Ta có:  $y=\sin 5x\cos 2x=\frac{1}{2}\left(\sin 7x+\sin 3x\right)$.

Đáp án B

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

Xét đáp án A,  $4y-{y^{\text{'}}}^{\text{'}}=4\sin 2x+4\sin 2x$.

Xét đáp án B,  $4y+{y^{\text{'}}}^{\text{'}}=4\sin 2x-4\sin 2x=0$.

Xét đáp án C,  $y^{\text{'}}\tan 2x=2\cos 2x.\frac{\sin 2x}{\cos 2x}=2\sin 2x\ne y$.

Xét đáp án D,  $y^2+{\left(y^{\text{'}}\right)}^2={\sin }^{2}2x+4{\cos }^{2}2x\ne 4$.

Đáp án C

Tập xác định:  $D=ℝ$.

Ta có  $y^{\text{'}}=3x^2-6x+1\Rightarrow {y^{\text{'}}}^{\text{'}}=6x-6\Rightarrow {y^{\text{'}}}^{\text{'}}=0\iff x=1$.

Đáp án D

Ta có:  $f^{\text{'}}\left(x\right)=4x^3+2\sin 2x$, suy ra  ${f^{\text{'}}}^{\text{'}}\left(x\right)=12x^2+4\cos 2x\Rightarrow {{f^{\text{'}}}^{\text{'}}}^{\text{'}}\left(x\right)=24x-8\sin 2x$.

Do đó:  $f^{\left(4\right)}\left(x\right)={\left({{f^{\text{'}}}^{\text{'}}}^{\text{'}}\left(x\right)\right)}^{\text{'}}=24-16\cos 2x$.

Đáp án D

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

Xét  $y.y^{\text{'}}=\sqrt{x^2+1}.\frac{x}{\sqrt{x^2+1}}=x$, do đó khẳng định (I) sai.

Xét   $y^2.{y^{\text{'}}}^{\text{'}}=\left(x^2+1\right).\frac{1}{\left(x^2+1\right)\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+1}}\ne y^{\text{'}}$, do đó khẳng định (II) sai.

Đáp án A

Ta có

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

Đáp án A

Ta có:

$f\left(x\right)=\sqrt{2x-1}\Rightarrow f^{\text{'}}\left(x\right)=\frac{{\left(2x-1\right)}^{\text{'}}}{2\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\Rightarrow {f^{\text{'}}}^{\text{'}}\left(x\right)=\frac{-{\left(\sqrt{2x-1}\right)}^{\text{'}}}{2x-1}=\frac{-1}{\left(2x-1\right)\sqrt{2x-1}}=\frac{-1}{\sqrt{{\left(2x-1\right)}^3}}$.

$\Rightarrow {{f^{\text{'}}}^{\text{'}}}^{\text{'}}\left(x\right)=\frac{{\left(\sqrt{{\left(2x-1\right)}^3}\right)}^{\text{'}}}{{\left(2x-1\right)}^3}=\frac{3{\left(2x-1\right)}^2}{{\left(2x-1\right)}^3\sqrt{{\left(2x-1\right)}^3}}=\frac{3}{\sqrt{{\left(2x-1\right)}^5}}$.

Vậy  ${{f^{\text{'}}}^{\text{'}}}^{\text{'}}\left(1\right)=3$.

Đáp án A

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

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

 $\Rightarrow {{f^{\text{'}}}^{\text{'}}}^{\text{'}}\left(x\right)=8\sin \left(2x-\frac{\pi }{3}\right)$

 $$$\Rightarrow f^{\left(4\right)}\left(x\right)=16\cos \left(2x-\frac{\pi }{3}\right)$.

Xét phương trình

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

Mà $x\in \left[0;\frac{\pi }{2}\right]$ nên chỉ có giá trị  $x=\frac{\pi }{2}$ thỏa mãn.

Đáp án D

Ta có  $y^{\text{'}}=-\frac{1}{x^2}\Rightarrow {y^{\text{'}}}^{\text{'}}=\frac{2}{x^3}$.

Xét đáp án A,  ${y^{\text{'}}}^{\text{'}}{y}^3+2=0\iff \frac{2}{x^3}.{\left(\frac{1}{x}\right)}^3+2=0\iff \frac{2}{x^6}+2=0$ (vô lí).

Xét đáp án B,  ${y^{\text{'}}}^{\text{'}}y+2{\left(y^{\text{'}}\right)}^2=0\iff \frac{2}{x^3}.\frac{1}{x}+2{\left(-\frac{1}{x^2}\right)}^2=0\iff \frac{4}{x^4}=0$ (vô lí).

Xét đáp án C,  ${y^{\text{'}}}^{\text{'}}{y}^3=2\iff \frac{2}{x^3}.{\left(\frac{1}{x}\right)}^3=2\iff \frac{2}{x^6}=2$ (vô lí).

Xét đáp án D,  ${y^{\text{'}}}^{\text{'}}y=2{\left(y^{\text{'}}\right)}^2\iff \frac{2}{x^3}.\frac{1}{x}=2{\left(-\frac{1}{x^2}\right)}^2\iff \frac{2}{x^4}=\frac{2}{x^4}$ (đúng).

Đáp án D

Ta có  $y^{\text{'}}=2\cos \left(2x+\frac{\pi }{2}\right);{y^{\text{'}}}^{\text{'}}=2^2\cos \left(2x+2\frac{\pi }{2}\right);{{y^{\text{'}}}^{\text{'}}}^{\text{'}}=2^3\cos \left(2x+3\frac{\pi }{2}\right)$.

Bằng quy nạp ta chứng minh được  $y^{\left(n\right)}=2^n\cos \left(2x+n\frac{\pi }{2}\right)$.

Đáp án D

Ta có  $y^{\text{'}}=\frac{1}{\sqrt{2x+1}},{y^{\text{'}}}^{\text{'}}=-\frac{1}{\sqrt{{\left(2x+1\right)}^3}},{{y^{\text{'}}}^{\text{'}}}^{\text{'}}=\frac{3}{\sqrt{{\left(2x+1\right)}^5}}$.

Bằng quy nạp ta chứng minh được  $y^{\left(n\right)}=\frac{{\left(-1\right)}^{n+1}\mathrm{.3.5...}\left(2n-3\right)}{\sqrt{{\left(2x+1\right)}^{2n-1}}}$.

Đáp án D

Ta có:  $y=\frac{5}{x-2}-\frac{3}{x-1}$.

Bằng quy nạp ta chứng minh được  $y^{\left(n\right)}=\frac{5.{\left(-1\right)}^n.n!}{{\left(x-2\right)}^{n+1}}-\frac{3.{\left(-1\right)}^n.n!}{{\left(x-1\right)}^{n+1}}$.

Đáp án D

Ta có:  $x=3\left(x+2\right)-2\left(x+3\right);x^2+5x+6=\left(x+2\right)\left(x+3\right)$.

Suy ra  $y=\frac{3}{x+3}-\frac{2}{x+2}$.

Mà  ${\left(\frac{1}{x+2}\right)}^{\left(n\right)}=\frac{{\left(-1\right)}^n{.1}^n.n!}{{\left(x+2\right)}^{n+1}}=\frac{{\left(-1\right)}^n.n!}{{\left(x+2\right)}^{n+1}},{\left(\frac{1}{x+3}\right)}^{\left(n\right)}=\frac{{\left(-1\right)}^n.n!}{{\left(x+3\right)}^{n+1}}$ 

nên ta có  $y^{\left(n\right)}=\frac{{\left(-1\right)}^n\mathrm{.3.}n!}{{\left(x+3\right)}^{n+1}}-\frac{{\left(-1\right)}^n\mathrm{.2.}n!}{{\left(x+2\right)}^{n+1}}$.

Đáp án C

Ta có  $f^{\text{'}}\left(x\right)=-\sin \left(x+a\right)=\cos \left(x+a+\frac{\pi }{2}\right);$

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

…

$f^{\left(2021\right)}\left(x\right)=\cos \left(x+a+\frac{2021\pi }{2}\right)=\cos \left(x+a+\frac{\pi }{2}\right)$.

Đáp án D

Ta có:  $y^{\text{'}}=2\sin \left(2x+\frac{\pi }{2}\right),{y^{\text{'}}}^{\text{'}}=2^2\sin \left(2x+2\frac{\pi }{2}\right),{{y^{\text{'}}}^{\text{'}}}^{\text{'}}=2^3\sin \left(2x+3\frac{\pi }{2}\right);\mathrm{...}$

Bằng phương pháp quy nạp ta chứng minh được  $y^{\left(n\right)}=2^n\sin \left(2x+n\frac{\pi }{2}\right)$.

Đáp án D

Ta có  $y=\sin 3x.\cos x-\sin 2x=\frac{1}{2}\left(\sin 4x+\sin 2x\right)-\sin 2x=\frac{1}{2}\left(\sin 4x-\sin 2x\right)$.

Mặt khác theo quy nạp ta chứng minh được  ${\left(\sin ax\right)}^{\left(n\right)}={\left(-1\right)}^{n-1}{a}^n\sin \left(\frac{n\pi }{2}-ax\right)$.

Do đó  $y^{\left(10\right)}\left(x\right)=\frac{1}{2}\left({\left(-1\right)}^9{4}^{10}.\sin \left(5\pi -4x\right)-{\left(-1\right)}^9{.2}^{10}.\sin \left(5\pi -2x\right)\right)=\frac{1}{2}\left(-4^{10}.\sin 4x+2^{10}\sin 2x\right)$

$\Rightarrow y^{\left(10\right)}\left(\frac{\pi }{3}\right)\approx \mathrm{454490,13}$

Đáp án A

Giả sử  $f\left(x\right)=a_0+a_{1}x+a_2{x}^2+\mathrm{...}+a_{18}{x}^{18}$.

Khi đó  $f^{\left(6\right)}\left(x\right)=6!.a_6+b_{7}x+b_8{x}^2+\mathrm{...}+b_{18}{x}^{12}\Rightarrow f^{\left(6\right)}\left(0\right)=720a_6$.

Ta có  ${\left(3x^2-2x-1\right)}^9=-{\left(1+2x-3x^2\right)}^9=-\sum _{k=0}^9{C}_9^k{\left(2x-3x^2\right)}^k$

$=-\sum _{k=0}^9{C}_9^k\sum _{i=0}^k{C}_k^i{\left(2x\right)}^{k-i}{\left(-3x^2\right)}^i=-\sum _{k=0}^9\sum _{i=0}^k{C}_9^k{C}_k^i{2}^{k-i}{\left(-3\right)}^i{x}^{k+i}$

Số hạng chứa $x^6$ ứng với $k,i$ thỏa mãn $\left\{\begin{array}{l}0\le i\le k\le 9\\ k+i=6\end{array}\right.\Rightarrow \left(k;i\right)\in \left\{\left(6;0\right),\left(5;1\right),\left(4;2\right),\left.\left(3;3\right)\right\}\right.$