Tìm vi phân của các hàm số \(y = {x^3} + 2{x^2}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\({\rm{d}}y = \left( { - \cos x + 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = \left( { - \cos x - 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = - \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

\[{\rm{d}}y = -\sin 2x\,{\rm{d}}x\]

\[{\rm{d}}y = \sin 2x\,{\rm{d}}x\]

\[{\rm{d}}y = \sin x\,{\rm{d}}x\]

\[{\rm{d}}y = {\rm{2cos}}x\,{\rm{d}}x\]

\[{\rm{d}}y = \left( {x\cos x-\sin x} \right){\rm{d}}x\]

\[{\rm{d}}y = \left( {x\cos x} \right){\rm{d}}x\]

\[{\rm{d}}y = \left( {\cos x-\sin x} \right){\rm{d}}x\].

\[{\rm{d}}y = \left( {x\sin x} \right){\rm{d}}x\]

\[dy = \frac{{1 - {x^2}}}{{{{({x^2} + 1)}^2}}}dx\]

\[dy = \frac{{2x}}{{({x^2} + 1)}}dx\]

\[dy = \frac{{1 - {x^2}}}{{({x^2} + 1)}}dx\]

\[dy = \frac{1}{{{{({x^2} + 1)}^2}}}dx\]

.$dy=2(x-1)dx$

$dy=2(x-1)$

$dy=(x-1)dx$

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