Câu hỏi
Tính đạo hàm của các hàm số sau:
Câu 1:
\(y = {\sin ^2}\left( {1 - x + {x^3}} \right)\)
- A \(\left( {3{x^2} - 1} \right)\sin \left( {2 - 2x + 2{x^3}} \right)\)
- B \(\left( {3{x^2} - 3x-1} \right)\sin \left( {2 +2x + 2{x^3}} \right)\)
- C \(\left( {{x^2} - 3x-1} \right)\cos \left( {2 +2x + 2{x^3}} \right)\)
- D \(\left( {2{x^2} - 3x -1} \right)\cos \left( {2 + 2{x^3}} \right)\)
Phương pháp giải:
Sử dụng công thức \(\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'\), \(\left( {\sin u} \right)' = u'.\cos u\).
Lời giải chi tiết:
\(y = {\sin ^2}\left( {1 - x + {x^3}} \right)\).
\(\begin{array}{l}y' = 2\sin \left( {1 - x + {x^3}} \right).\left[ {\sin \left( {1 - x + {x^3}} \right)} \right]'\\y' = 2\sin \left( {1 - x + {x^3}} \right).\left( {1 - x + {x^3}} \right)'.\cos \left( {1 - x + {x^3}} \right)\\y' = \left( {3{x^2} - 1} \right)\sin \left( {2 - 2x + 2{x^3}} \right).\end{array}\)
Chọn A.
Câu 2:
\(y = {\cos ^3}x\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\)
- A \(\dfrac{{ - 21}}{{{{\left( {x -1 } \right)}^2}}}.{\cos ^2}\left( {\dfrac{{2x + 1}}{{x-3}}} \right).sin\left( {\dfrac{{2x + 3}}{{3 - x}}} \right)\)
- B \(\dfrac{{ - 21}}{{{{\left( {x -3} \right)}^2}}}.{\sin^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right).sin\left( {\dfrac{{2x + 3}}{{1 - x}}} \right)\)
- C \(\dfrac{{ - 21}}{{{{\left( {3 - x} \right)}^2}}}.{\cos ^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right).sin\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\)
- D \(\dfrac{{ - 21}}{{{{\left( {1 - x} \right)}^2}}}.{\sin^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right).cos\left( {\dfrac{{2x + 1}}{{1 - x}}} \right)\)
Phương pháp giải:
Sử dụng công thức \(\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'\), \(\left( {\cos u} \right)' = - u'\sin u\).
Lời giải chi tiết:
\(y = {\cos ^3}x\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\)
\(\begin{array}{l}y' = 3{\cos ^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\left[ {\cos \left( {\dfrac{{2x + 1}}{{3 - x}}} \right)} \right]'\\y' = - 3{\cos ^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\left[ {\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)} \right]'sin\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\\y' = - 3{\cos ^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right).\dfrac{{2\left( {3 - x} \right) + \left( {2x + 1} \right)}}{{{{\left( {3 - x} \right)}^2}}}.sin\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\\y' = \dfrac{{ - 21}}{{{{\left( {3 - x} \right)}^2}}}.{\cos ^2}\left( {\dfrac{{2x + 1}}{{3 - x}}} \right).sin\left( {\dfrac{{2x + 1}}{{3 - x}}} \right)\end{array}\).
Chọn C.
Câu 3:
\(y = {\tan ^5}\left( {\cos \sqrt x } \right)\)
- A \(\dfrac{{ {{\tan }^3}\left( {\cos \sqrt x } \right).\sin \sqrt x }}{{4\sqrt x .{{\cos }^2}\left( {\sin\ x } \right)}}\)
- B \(\dfrac{{ 5{{\tan }^3}\left( {\cos \sqrt x } \right).\sin \sqrt x }}{{4\sqrt x .{{\cos }^2}\left( {\cos \sqrt x } \right)}}\)
- C \(\dfrac{{ - 5{{\tan }^4}\left( {\cos \sqrt x } \right).\sin \sqrt x }}{{2\sqrt x .{{\cos }^2}\left( {\cos \sqrt x } \right)}}\)
- D \(\dfrac{{ - 5{{\tan }^2}\left( {\sin ^2\sqrt x } \right) }}{{2\sqrt x .{{\cos }^4}\left( {\cos \sqrt x } \right)}}\)
Phương pháp giải:
Sử dụng công thức \(\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'\), \(\left( {\tan u} \right)' = \dfrac{{u'}}{{{{\cos }^2}u}}\).
Lời giải chi tiết:
\(y = {\tan ^5}\left( {\cos \sqrt x } \right)\)
\(\begin{array}{l}y' = 5{\tan ^4}\left( {\cos \sqrt x } \right)\left[ {\tan \left( {\cos \sqrt x } \right)} \right]'\\y' = 5{\tan ^4}\left( {\cos \sqrt x } \right)\dfrac{{\left( {\cos \sqrt x } \right)'}}{{{{\cos }^2}\left( {\cos \sqrt x } \right)}}\\y' = - 5{\tan ^4}\left( {\cos \sqrt x } \right)\dfrac{{\left( {\sqrt x } \right)'\left( {\sin \sqrt x } \right)}}{{{{\cos }^2}\left( {\cos \sqrt x } \right)}}\\y' = - 5{\tan ^4}\left( {\cos \sqrt x } \right)\dfrac{{\sin \sqrt x }}{{2\sqrt x .{{\cos }^2}\left( {\cos \sqrt x } \right)}}\\y' = \dfrac{{ - 5{{\tan }^4}\left( {\cos \sqrt x } \right).\sin \sqrt x }}{{2\sqrt x .{{\cos }^2}\left( {\cos \sqrt x } \right)}}\end{array}\).
Chọn C.
Câu 4:
\(y = {\cot ^8}\left( {{{\sin }^2}x} \right)\)
- A \(\dfrac{{ 8{{\cos }^5}.\sin 2x}}{{{{\sin }^7}\left( {{{\sin }^4}x} \right)}}\)
- B \(\dfrac{{ - 8{{\cos }^7}\left( {{{\sin }^2}x} \right).\sin 2x}}{{{{\sin }^9}\left( {{{\sin }^2}x} \right)}}\)
- C \(\dfrac{{ 8{{\cos }^7}\left( {{{\sin }^2}x} \right).\sin 2x}}{{{{\sin }^5}\left( {{{\sin }^2}x} \right)}}\)
- D \(\dfrac{{ - 8{{\cos }^7}\left( {{{\sin }}x} \right).\sin 2x}}{{{{\sin }^9}\left( {{{\cos}^2}x} \right)}}\)
Phương pháp giải:
Sử dụng công thức \(\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'\), \(\left( {\cot u} \right)' = \dfrac{{ - u'}}{{{{\sin }^2}u}}\).
Lời giải chi tiết:
\(y = {\cot ^8}\left( {{{\sin }^2}x} \right)\)
\(\begin{array}{l}y' = 8{\cot ^7}\left( {{{\sin }^2}x} \right).\left[ {\cot \left( {{{\sin }^2}x} \right)} \right]'\\y' = 8{\cot ^7}\left( {{{\sin }^2}x} \right).\dfrac{{ - \left( {{{\sin }^2}x} \right)'}}{{{{\sin }^2}\left( {{{\sin }^2}x} \right)}}\\y' = 8{\cot ^7}\left( {{{\sin }^2}x} \right).\dfrac{{ - 2\sin x\left( {\sin x} \right)'}}{{{{\sin }^2}\left( {{{\sin }^2}x} \right)}}\\y' = - 8{\cot ^7}\left( {{{\sin }^2}x} \right).\dfrac{{\sin 2x}}{{{{\sin }^2}\left( {{{\sin }^2}x} \right)}}\\y' = - 8\dfrac{{{{\cos }^7}\left( {{{\sin }^2}x} \right)}}{{{{\sin }^7}\left( {{{\sin }^2}x} \right)}}.\dfrac{{\sin 2x}}{{{{\sin }^2}\left( {{{\sin }^2}x} \right)}}\\y' = \dfrac{{ - 8{{\cos }^7}\left( {{{\sin }^2}x} \right).\sin 2x}}{{{{\sin }^9}\left( {{{\sin }^2}x} \right)}}\end{array}\)
Chọn B.
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