Tính đạo hàm a) \(y = {\left( {\dfrac{{{x^2} + 1}}{x}} \right)^3}\)b) \(y = {\sin ^2}2x\)c) \(y = {\cos ^4}\left(

Câu hỏi số 620714:

Vận dụng

Tính đạo hàm

a) \(y = {\left( {\dfrac{{{x^2} + 1}}{x}} \right)^3}\)

b) \(y = {\sin ^2}2x\)

c) \(y = {\cos ^4}\left( {\sqrt x } \right)\)

d) \(y = \sqrt {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} \)

e) \(y = \sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \)

g) \(y = \sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} \)

h) \(y = \sqrt {\sin 4x + x\cos x} \)

i) \(y = \sqrt {{{\cos }^4}\left( { - 3x} \right)} \)

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Câu hỏi:620714

Giải chi tiết

a) \(y = {\left( {\dfrac{{{x^2} + 1}}{x}} \right)^3}\)

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

b) \(y = {\sin ^2}2x\)

\(y' = 2\sin 2x.\left( {\sin 2x} \right)' = 2\sin 2x.\cos 2x.\left( {2x} \right)' = 4\sin 2x\cos 2x\)

c) \(y = {\cos ^4}\left( {\sqrt x } \right)\)

\(\begin{array}{l}y' = 4{\cos ^3}\left( {\sqrt x } \right).\left[ {\cos \left( {\sqrt x } \right)} \right]'\\y' = 4{\cos ^3}\left( {\sqrt x } \right).\left( { - \sin \left( {\sqrt x } \right)} \right).\left( {\sqrt x } \right)'\\y' = - 4\sin \left( {\sqrt x } \right){\cos ^3}\left( {\sqrt x } \right).\dfrac{1}{{2\sqrt x }}\\y' = \dfrac{{ - 2\sin \left( {\sqrt x } \right){{\cos }^3}\left( {\sqrt x } \right)}}{{\sqrt x }}\end{array}\)

d) \(y = \sqrt {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} \)

\(\begin{array}{l}y' = \dfrac{1}{{2\sqrt {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} }}.\left( {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} \right)'\\y' = \dfrac{1}{{2\sqrt {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} }}.\dfrac{{\left( {2x + 2} \right)\left( {x + 3} \right) - \left( {{x^2} + 2x - 1} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\y' = \dfrac{1}{{2\sqrt {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} }}.\dfrac{{2{x^2} + 8x + 6 - {x^2} - 2x + 1}}{{{{\left( {x + 3} \right)}^2}}}\\y' = \dfrac{1}{{2\sqrt {\dfrac{{{x^2} + 2x - 1}}{{x + 3}}} }}.\dfrac{{{x^2} + 6x + 7}}{{{{\left( {x + 3} \right)}^2}}}\end{array}\)

e) \(y = \sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \)

\(\begin{array}{l}y' = \dfrac{1}{{2\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}.\left( {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right)'\\y' = \dfrac{1}{{2\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}.\dfrac{{\left( {\cos x - \sin x} \right)\left( {\sin x - \cos x} \right) - \left( {\sin x + \cos x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\y' = \dfrac{1}{{2\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}.\dfrac{{ - {{\left( {\cos x - \sin x} \right)}^2} - {{\left( {\sin x + \cos x} \right)}^2}}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\y' = \dfrac{1}{{2\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}.\dfrac{{ - {{\cos }^2}x + 2\sin x\cos x - {{\sin }^2}x - {{\sin }^2}x + 2\sin x\cos x - {{\cos }^2}x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\y' = \dfrac{1}{{2\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}.\dfrac{{ - 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\y' = \dfrac{1}{{2\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}.\dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\y' = \dfrac{{ - 1}}{{{{\left( {\sin x - \cos x} \right)}^2}\sqrt {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} }}\end{array}\)

g) \(y = \sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} \)

\(\begin{array}{l}y' = \dfrac{1}{{2\sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} }}.\left( {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} \right)'\\y' = \dfrac{1}{{2\sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} }}.4{\sin ^3}\left( {\dfrac{{x + 2}}{{x + 1}}} \right).\left( {\sin \left( {\dfrac{{x + 2}}{{x + 1}}} \right)} \right)'\\y' = \dfrac{1}{{\sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} }}.2{\sin ^3}\left( {\dfrac{{x + 2}}{{x + 1}}} \right).\cos \left( {\dfrac{{x + 2}}{{x + 1}}} \right).\left( {\dfrac{{x + 2}}{{x + 1}}} \right)'\\y' = \dfrac{1}{{\sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} }}.2{\sin ^3}\left( {\dfrac{{x + 2}}{{x + 1}}} \right).\cos \left( {\dfrac{{x + 2}}{{x + 1}}} \right).\dfrac{{\left( {x + 1} \right) - \left( {x + 2} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\y' = \dfrac{1}{{\sqrt {{{\sin }^4}\left( {\dfrac{{x + 2}}{{x + 1}}} \right)} }}.2{\sin ^3}\left( {\dfrac{{x + 2}}{{x + 1}}} \right).\cos \left( {\dfrac{{x + 2}}{{x + 1}}} \right).\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}\\\end{array}\)

h) \(y = \sqrt {\sin 4x + x\cos x} \)

\(\begin{array}{l}y' = \dfrac{1}{{2\sqrt {\sin 4x + x\cos x} }}.\left( {\sin 4x + x\cos x} \right)'\\y' = \dfrac{1}{{2\sqrt {\sin 4x + x\cos x} }}.\left( {4\cos 4x + \cos x - x\sin x} \right)\end{array}\)

i) \(y = \sqrt {{{\cos }^4}\left( { - 3x} \right)} \)

\(\begin{array}{l}y' = \dfrac{1}{{2\sqrt {{{\cos }^4}\left( { - 3x} \right)} }}\left( {{{\cos }^4}\left( { - 3x} \right)} \right)'\\y' = \dfrac{1}{{2\sqrt {{{\cos }^4}\left( { - 3x} \right)} }}4{\cos ^3}\left( { - 3x} \right)\left( {\cos \left( { - 3x} \right)} \right)'\\y' = \dfrac{1}{{2\sqrt {{{\cos }^4}\left( { - 3x} \right)} }}4{\cos ^3}\left( { - 3x} \right)\left( { - \sin \left( { - 3x} \right)} \right)\left( { - 3x} \right)'\\y' = \dfrac{1}{{2\sqrt {{{\cos }^4}\left( { - 3x} \right)} }}4{\cos ^3}\left( { - 3x} \right)\left( { - \sin \left( { - 3x} \right)} \right)\left( { - 3} \right)\\y' = \dfrac{{6\sin \left( { - 3x} \right){{\cos }^3}\left( { - 3x} \right)}}{{{{\cos }^2}\left( { - 3x} \right)}}\\y' = 6\sin \left( { - 3x} \right)\cos \left( { - 3x} \right)\\y' = 3\sin \left( { - 6x} \right)\end{array}\)

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