ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Các quy tắc tính đạo hàm

Bước 1:

\[\begin{array}{*{20}{l}}{y = \sin 2x = 2\sin x\cos x}\\{ \Rightarrow y' = {{\left( {2\sin x\cos x} \right)}^\prime }}\\{ = 2{{\left( {\sin x\cos x} \right)}^\prime }}\\{ = 2\left[ {{{\left( {\sin x} \right)}^\prime }.\cos x + \sin x.{{\left( {\cos x} \right)}^\prime }} \right]}\end{array}\]

Bước 2:

\[ = 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right) = 2\cos 2x\]

\[y\prime = \frac{{(2{x^2} + 3x - 1)\prime ({x^2} - 5x + 2) - (2{x^2} + 3x - 1)({x^2} - 5x + 2)\prime }}{{{{({x^2} - 5x + 2)}^2}}}\]

\[y\prime = \frac{{(4x + 3)({x^2} - 5x + 2) - (2{x^2} + 3x - 1)(2x - 5)}}{{{{({x^2} - 5x + 2)}^2}}}\]

\( = \frac{{4{x^3} - 20{x^2} + 8x + 3{x^2} - 15x + 6 - 4{x^3} - 6{x^2} + 2x + 10{x^2} + 15x - 5}}{{{{({x^2} - 5x + 2)}^2}}}\)

\[y\prime = \frac{{ - 13{x^2} + 10x + 1}}{{{{({x^2} - 5x + 2)}^2}}}\]

\[f(x) = {(\sqrt x - \frac{1}{{\sqrt x }})^3} = {(\sqrt x )^3} - 3{(\sqrt x )^2}.\frac{1}{{\sqrt x }} + 3\sqrt x {(\frac{1}{{\sqrt x }})^2} - {(\frac{1}{{\sqrt x }})^3}\]

\[f(x) = {x^{\frac{3}{2}}} - 3\sqrt x + \frac{3}{{\sqrt x }} - \frac{1}{{{x^{\frac{3}{2}}}}}\]

\[f(x) = {x^{\frac{3}{2}}} - 3\sqrt x + 3{x^{ - \frac{1}{2}}} - {x^{ - \frac{3}{2}}}\]

\[f\prime (x) = \frac{3}{2}{x^{\frac{3}{2} - 1}} - \frac{3}{{2\sqrt x }} + 3.( - \frac{1}{2}){x^{ - \frac{1}{2} - 1}} + \frac{3}{2}{x^{ - \frac{3}{2} - 1}}\]

\[f\prime (x) = \frac{3}{2}\sqrt x - \frac{3}{{2\sqrt x }} - \frac{3}{2}{x^{ - \frac{3}{2}}} + \frac{3}{2}{x^{ - \frac{5}{3}}}\]

\[f\prime (x) = \frac{3}{2}(\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }})\]

\[y = ta{n^2}x - co{t^2}x = (tanx - cotx)(tanx + cotx)\]

\[y\prime = (tanx - cotx)\prime (tanx + cotx) + (tanx - cotx)(tanx + cotx)\prime \]

\[y\prime = (\frac{1}{{co{s^2}x}} + \frac{1}{{si{n^2}x}})(tanx + cotx) + (tanx - cotx)(\frac{1}{{co{s^2}x}} - \frac{1}{{si{n^2}x}})\]

\[y\prime = \frac{{tanx}}{{co{s^2}x}} + \frac{{cotx}}{{co{s^2}x}} + \frac{{tanx}}{{si{n^2}x}} + \frac{{cotx}}{{si{n^2}x}} + \frac{{tanx}}{{co{s^2}x}} - \frac{{tanx}}{{si{n^2}x}} - \frac{{cotx}}{{co{s^2}x}} + \frac{{cotx}}{{si{n^2}x}}\]

\[y\prime = 2\frac{{tanx}}{{co{s^2}x}} + 2\frac{{cotx}}{{si{n^2}x}}\]

\[f(x) = tan(x - \frac{{2\pi }}{3}) = \frac{{tanx - tan\frac{{2\pi }}{3}}}{{1 + tanx.tan\frac{{2\pi }}{3}}} = \frac{{tanx + \sqrt 3 }}{{1 - \sqrt 3 tanx}}.(tanx + \sqrt 3 )\prime (1 - \sqrt 3 tanx)\]

\[f\prime (x) = \frac{{ - (tanx + \sqrt 3 )(1 - \sqrt 3 tanx)\prime }}{{{{\left( {1 - \sqrt 3 tanx} \right)}^2}}}\]

\[f\prime (x) = \frac{{\frac{1}{{co{s^2}x}}(1 - \sqrt 3 tanx) - (tanx + \sqrt 3 )( - \frac{{\sqrt 3 }}{{co{s^2}x}})}}{{{{(1 - \sqrt 3 tanx)}^2}}}\]

\[f\prime (x) = \frac{{\frac{1}{{co{s^2}x}} - \frac{{\sqrt 3 tanx}}{{co{s^2}x}} + \frac{{\sqrt 3 tanx}}{{co{s^2}x}} + \frac{3}{{co{s^2}x}}}}{{{{(1 - \sqrt 3 tanx)}^2}}}\]

\(f\prime (x) = \frac{4}{{co{s^2}x{{(1 - \sqrt 3 tanx)}^2}}}\)

\[ \Rightarrow f\prime (0) = \frac{4}{{1\left( {1 - \sqrt 3 .0} \right)}} = 4\]

Bước 1:

\[\begin{array}{*{20}{l}}{{{\left( {{{\tan }^2}\frac{x}{2}} \right)}^\prime } = 2\tan \frac{x}{2}{{\left( {\tan \frac{x}{2}} \right)}^\prime }}\end{array}\]

Bước 2:

\[ = 2\tan \frac{x}{2}.\frac{{{{\left( {\frac{x}{2}} \right)}^\prime }}}{{{{\cos }^2}\frac{x}{2}}}\]

\[ = 2\tan \frac{x}{2}.\frac{{\frac{1}{2}}}{{{{\cos }^2}\frac{x}{2}}} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}.\frac{1}{{{{\cos }^2}\frac{x}{2}}} = \frac{{\sin \frac{x}{2}}}{{{{\cos }^3}\frac{x}{2}}}\]

\[y = x(2x - 1)(3x + 2)(sinx - cosx)\prime \]

\[ = (6{x^3} + {x^2} - 2x)(sinx + cosx)\]

\[ \Rightarrow y\prime = (6{x^3} + {x^2} - 2x)\prime (sinx + cosx) + (6{x^3} + {x^2} - 2x)(sinx + cosx)\prime \]

\[y\prime = (18{x^2} + 2x - 2)(sinx + cosx) + (6{x^3} + {x^2} - 2x)(cosx - sinx)\]

\[y\prime = sinx(18{x^2} + 2x - 2 - 6{x^3} - {x^2} + 2x) + cosx(18{x^2} + 2x - 2 + 6{x^3} + {x^2} - 2x)\]

\[y\prime = sinx( - 6{x^3} + 17{x^2} + 4x - 2) + cosx(6{x^3} + 19{x^2} - 2)\]

Với x>1 ta có:\[f\left( x \right) = {x^2} - 3x + 1 \Rightarrow f'\left( x \right) = 2x - 3\]

Với x<1 ta có : \[f\left( x \right) = 2x + 2 \Leftrightarrow f'\left( x \right) = 2\]

Với x=1 ta có :\[\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - 3x + 1} \right) = - 1 \ne f\left( 1 \right) = 4\] Hàm số không liên tục tại x=1, do đó không có đạo hàm tại x=1.

Vậy\(f'\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{2x - 3\,khi\,x > 1}\\{2\,khi\,x < 1}\end{array}} \right.\)

Đáp án cần chọn là: B

\[\begin{array}{*{20}{l}}{y = \frac{{m{x^3}}}{3} - m{x^2} + \left( {3m - 1} \right)x + 1}\\{ \Rightarrow y' = m{x^2} - 2mx + 3m - 1}\\{y' \le 0,\forall x \in R \Rightarrow m{x^2} - 2mx + 3m - 1 \le 0,\forall x \in R}\end{array}\]

TH1: \[m = 0,\] khi đó \[BPT \Leftrightarrow - 1 \le 0\] đúng\[\forall x \in R\]

TH2:

\[m \ne 0 \Leftrightarrow y\prime \le 0\forall x \in R \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = m < 0}\\{\Delta \prime = {m^2} - m(3m - 1) \le 0}\end{array}} \right.\]

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{m < 0}\\{ - 2{m^2} + m \le 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{m < 0}\\{\left[ {\begin{array}{*{20}{c}}{m \le 0}\\{m \ge \frac{1}{2}}\end{array}} \right.}\end{array}} \right.\)

\[ \Leftrightarrow m < 0\]

Kết hợp cả 2 trường hợp ta có\[m \le 0\] là những giá trị cần tìm.

Bước 1:

\[\frac{1}{{{x^2}}} = {x^{ - 2}}\]

Bước 2:

\[ \Rightarrow y' = {\left( {{x^{ - 2}}} \right)^\prime } = - 2.{x^{ - 3}} = \frac{{ - 2}}{{{x^3}}}\]

\[y' = {\left( {2\sin x - 3\cos x} \right)^\prime }\]

\[ = {\left( {2\sin x} \right)^\prime } - {\left( {3\cos x} \right)^\prime }\]

\[ = 2(\sin x)' - 3(\cos x)'\]

\[ = 2.\cos x + 3\sin x\]

Bước 1:

\[y' = {\left( {3x - 1} \right)^\prime }.\sqrt {{x^2} + 1} + \left( {3x - 1} \right).{\left( {\sqrt {{x^2} + 1} } \right)^\prime }\]

Bước 2:

\[\begin{array}{*{20}{l}}{ = 3.\sqrt {{x^2} + 1} + \left( {3x - 1} \right).\frac{{{{\left( {{x^2} + 1} \right)}^\prime }}}{{2.\sqrt {{x^2} + 1} }}}\\{ = 3.\sqrt {{x^2} + 1} + \left( {3x - 1} \right).\frac{{2x}}{{2.\sqrt {{x^2} + 1} }}}\\{ = 3.\sqrt {{x^2} + 1} + \left( {3x - 1} \right).\frac{x}{{\sqrt {{x^2} + 1} }}}\end{array}\]

Bước 3:

\[\begin{array}{*{20}{l}}{ = \frac{{3.\left( {{x^2} + 1} \right) + 3{x^2} - x}}{{\sqrt {{x^2} + 1} }}}\\{ = \frac{{6{x^2} - x + 3}}{{\sqrt {{x^2} + 1} }}}\end{array}\]

Đáp án cần chọn là: D

Bước 1:

\[\begin{array}{*{20}{l}}{y' = \frac{{{{\left( {10x - {x^2}} \right)}^\prime }}}{{2\sqrt {10x - {x^2}} }} = \frac{{10 - 2x}}{{2\sqrt {10x - {x^2}} }}}\\{ = \frac{{5 - x}}{{\sqrt {10x - {x^2}} }}}\end{array}\]

Bước 2:

Thay x=2 vào y′:

\[y'(2) = \frac{{5 - 2}}{{\sqrt {10 \cdot 2 - {2^2}} }} = \frac{3}{4}\]

Đáp án cần chọn là: C

Bước 1:

\[y' = \frac{{{{\left( {\sin 2x + 2} \right)}^\prime }\left( {\cos 2x + 3} \right) - {{\left( {\cos 2x + 3} \right)}^\prime }\left( {\sin 2x + 2} \right)}}{{{{\left( {\cos 2x + 3} \right)}^2}}}\]

Bước 2:

\[\begin{array}{*{20}{l}}{y' = \frac{{2cox2x\left( {\cos 2x + 3} \right) + 2\sin 2x\left( {\sin 2x + 2} \right)}}{{{{\left( {\cos 2x + 3} \right)}^2}}}}\\{ = \frac{{2\left( {{{\cos }^2}2x + {{\sin }^2}2x} \right) + 6\cos 2x + 4\sin 2x}}{{{{\left( {\cos 2x + 3} \right)}^2}}}}\\{ = \frac{{2\left( {3\cos 2x + 2\sin 2x + 1} \right)}}{{{{\left( {\cos 2x + 3} \right)}^2}}}}\end{array}\]

\[f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 2018} \right)\]

\[\begin{array}{*{20}{l}}{ \Rightarrow f'\left( x \right) = 1.\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 2018} \right) + x.1.\left( {x - 2} \right)...\left( {x - 2018} \right) + x\left( {x - 1} \right).1.\left( {x - 2} \right)...\left( {x - 2018} \right) + ... + }\\{x.\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 2017} \right).1}\end{array}\]\[ \Rightarrow f'\left( 0 \right) = 1.\left( { - 1} \right)\left( { - 2} \right)...\left( { - 2018} \right) + 0 + 0 + ... + 0 = 1.2...2018.{( - 1)^{2018}} = 2018!\]