Trắc nghiệm Toán 10 Bài 3 Đạo hàm của hàm số lượng giác

A. \(\frac{{\cos x + \sin x}}{{\sin x - \cos x}}\)

B. \(\frac{{ - 1}}{{{{\cos }^2}\left( {x + \frac{\pi }{4}} \right)}}\)

C. \(\frac{{ 1}}{{{{\cos }^2}\left( {x + \frac{\pi }{4}} \right)}}\)

D. \( - \frac{{2\sin x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\)

A. \(\frac{{3{{\sin }^2}\left( {2x + 1} \right)}}{{2\sqrt {{{\sin }^3}\left( {2x + 1} \right)} }}\)

B. \(3\sqrt {\sin \left( {2x + 1} \right)} .c{\rm{os}}\left( {2x + 1} \right)\)

C. \(\frac{{3{{\sin }^2}2\left( {2x + 1} \right)}}{{2\sqrt {{{\sin }^3}\left( {2x + 1} \right)} }}\)

D. \(3\sqrt {\sin \left( {2x + 1} \right)} \)

A. \(\frac{{15}}{{{{\left( {x - 2} \right)}^2}}}{\cos ^4}\frac{{x + 1}}{{x - 2}}\sin \frac{{x + 1}}{{x - 2}}\)

B. \(- 5{\cos ^4}\frac{{x + 1}}{{x - 2}}\sin \frac{{x + 1}}{{x - 2}}\)

C. \( 5{\cos ^4}\frac{{x + 1}}{{x - 2}}\sin \frac{{x + 1}}{{x - 2}}\)

D. \(\frac{{-15}}{{{{\left( {x - 2} \right)}^2}}}{\cos ^4}\frac{{x + 1}}{{x - 2}}\sin \frac{{x + 1}}{{x - 2}}\)

A. \(\frac{1}{{{{\sin }^2}x.\sqrt {3 + 2\tan x} }}\)

B. \(\frac{-1}{{{{\sin }^2}x.\sqrt {3 + 2\tan x} }}\)

C. \(\frac{1}{{{\rm{co}}{{\rm{s}}^2}x.\sqrt {3 + 2\tan x} }}\)

D. \(\frac{1}{{\cos x.\sqrt {3 + 2\tan x} }}\)

A.  -6cos⁵ xsinx

B. 6cos⁵ xsinx

C.  6sin⁵ xcosx

D. 6cos⁵ x

A. \(\frac{{2x - 1}}{{{{\sin }^2}\left( {{x^2} - x + 1} \right)}}\)

B. \(-\frac{{2x - 1}}{{\sin \left( {{x^2} - x + 1} \right)}}\)

C. \(\frac{{2x - 1}}{{{\rm{co}}{{\rm{s}}^2}\left( {{x^2} - x + 1} \right)}}\)

D. \(\frac{{1 - 2x}}{{{{\sin }^2}\left( {{x^2} - x + 1} \right)}}\)

A. \(\left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x =  - \frac{{3\pi }}{8} + k\frac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)\)

B. \(\begin{array}{l}
f'(x) = 0\\
\left[ \begin{array}{l}
x =  - \frac{\pi }{{12}} + k\pi \\
x =  - \frac{{3\pi }}{8} + k\frac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

C. \(\begin{array}{l}
f'(x) = 0\\
\left[ \begin{array}{l}
x =  - \frac{\pi }{{12}} + k\pi \\
x = \frac{{3\pi }}{8} + k\frac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

D. \(\begin{array}{l}
f'(x) = 0\\
\left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = \frac{{3\pi }}{8} + k\frac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

A. \(\cos 2x - \sin \frac{{{x^2} + 1}}{2} - \frac{1}{{{{\cos }^2}\sqrt x }}\)

B. \(2\cos 2x - x\sin \frac{{{x^2} + 1}}{2} - \frac{1}{{2\sqrt x .{{\cos }^2}\sqrt x }}\)

C. \( - 2\cos x + x\sin \frac{{{x^2} + 1}}{2} - \frac{1}{{2\sqrt x .{{\cos }^2}\sqrt x }}\)

D. \(2\cos 2x + x\sin \frac{{{x^2} + 1}}{2} - \frac{1}{{2\sqrt x .{{\cos }^2}\sqrt x }}\)

A. \(\cos 2x.4{\cos ^2}x + \frac{1}{{{{\sin }^2}\frac{1}{{{x^2}}}}} - \cos 2x.4{\sin ^3}x\)

B. \(2\cos 4x + \frac{2}{{{x^3}.{{\sin }^2}\frac{1}{{{x^2}}}}}\)

C. \(2\cos 4x + \frac{2}{{x.{{\sin }^2}\frac{1}{{{x^2}}}}}\)

D. \(2\cos 4x - \frac{2}{{{x^3}.{{\sin }^2}\frac{1}{{{x^2}}}}}\)

A. \(\frac{8}{9}\)

B. \(-\frac{9}{8}\)

C. \(\frac{9}{8}\)

D. \(-\frac{8}{9}\)