Tính đạo hàm cấp \(n\) của hàm số \(y = \frac{x}{{{x^2} + 5x + 6}}\)

\({y^{(n)}} = {2^{n + 1}}\cos \left( {2x + n\frac{\pi }{2}} \right)\)

\({y^{(n)}} = {2^{n - 1}}\cos \left( {2x + n\frac{\pi }{2}} \right)\)

\({y^{(n)}} = {2^n}\cos \left( {2x + \frac{\pi }{2}} \right)\)

\({y^{(n)}} = {2^n}\cos \left( {2x + n\frac{\pi }{2}} \right)\)

$\sum _{n=1}^{\infty }\frac{n}{n^4+1}$

$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}$

$\sum _{n=1}^{\infty }\frac{{(-1)}^n\cos \left(\mathrm{\pi n}\right)}{\sqrt{n}}$

$\sum _{n=1}^{\infty }\frac{1}{3^n}$

$A.\sum _{n=1}^{\infty }\frac{1}{n}$

$B.\sum _{n=1}^{\infty }{\left(\frac{1}{5}\right)}^n$

$C.\sum _{n=1}^{\infty }\left(\frac{1}{n^4}\right)$

$\underset{n=1}{\overset{\infty }{D.\sum }}\frac{1}{\sqrt{n}}$

$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^4+1}$

$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{3^n}$

\({\rm{d}}y = 2\left( {x - 1} \right){\rm{d}}x\)

\({\rm{d}}y = {\left( {x - 1} \right)^2}{\rm{d}}x\)

\({\rm{d}}y = 2\left( {x - 1} \right)\)

\({\rm{d}}y = 2\left( {x - 1} \right){\rm{d}}x\)

\(dy = (3{x^2} - 4x)dx\)

\(dy = (3{x^2} + x)dx\)

\(dy = (3{x^2} + 2x)dx\)

\(dy = (3{x^2} + 4x)dx\)

\(dy = \frac{3}{{\sqrt {3x + 2} }}dx\)

\(dy = \frac{1}{{2\sqrt {3x + 2} }}dx\)

\(dy = \frac{1}{{\sqrt {3x + 2} }}dx\)

\(dy = \frac{3}{{2\sqrt {3x + 2} }}dx\)

\(dy = (1 + {\tan ^2}2x)dx\)

\(dy = (1 - {\tan ^2}2x)dx\)

\(dy = 2(1 - {\tan ^2}2x)dx\)

\(dy = 2(1 + {\tan ^2}2x)dx\)

\({\rm{d}}f(x) = \frac{{ - \sin 4x}}{{2\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}f(x) = \frac{{ - \sin 4x}}{{\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}f(x) = \frac{{\cos 2x}}{{\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}f(x) = \frac{{ - \sin 2x}}{{2\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

\({\rm{d}}y = \frac{{{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = \frac{{3{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = \frac{{ - 3{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

\({\rm{d}}y = - \frac{{{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)