Tìm vi phân của các hàm số \(y = \sqrt[3]{{x + 1}}\)

\({\rm{d}}y = \left( {3{x^2} - 5} \right){\rm{d}}x\)

\({\rm{d}}y = - \left( {3{x^2} - 5} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {3{x^2} + 5} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {3{x^2} - 5} \right){\rm{d}}x\)

\({\rm{d}}y = \frac{1}{4}{\rm{d}}x\)

\({\rm{d}}y = \frac{1}{{{x^4}}}{\rm{d}}x\)

\({\rm{d}}y = - \frac{1}{{{x^4}}}{\rm{d}}x\)

\({\rm{d}}y = {x^4}{\rm{d}}x\)

\[{\rm{d}}y = \frac{{2\sqrt x }}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\]

\[{\rm{d}}y = \frac{{\sin (2\sqrt x )}}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\]

\[{\rm{d}}y = \frac{{2\sqrt x - \sin (2\sqrt x )}}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\]

\[{\rm{d}}y = - \frac{{2\sqrt x - \sin (2\sqrt x )}}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\]

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$dy=-2017\sin \left(2017x\right)dx$

$dy=\frac{2017}{{\sin }^2\left(2017x\right)}dx$

$dy=\frac{-2017}{co{s}^2\left(2017x\right)}dx$

$dy=\frac{-2017}{{\sin }^2\left(2017x\right)}dx$

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

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

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

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

$dy=\frac{5x}{{\cos }^{2}5x}$$dx$

$dy=\frac{-5x}{{\sin }^{2}5x}$$dx$

$dy=\frac{5}{co{s}^{2}5x}$$dx$

$dy=4\cos 2x\sin 2xdx$

$dy=2\cos 2x\sin 2xdx$

$dy=-2\cos 2x\sin 2xdx$

$dy=-2\sin 4xdx$

$df\left(x\right)=\frac{-\sin \left(4x\right)}{2\sqrt{1+{\cos }^{2}2x}}dx$

$df\left(x\right)=\frac{-\sin \left(4x\right)}{\sqrt{1+{\cos }^{2}2x}}dx$

$df\left(x\right)=\frac{\cos \left(2x\right)}{\sqrt{1+{\cos }^{2}2x}}dx$

$df\left(x\right)=\frac{-\sin \left(2x\right)}{\sqrt{1+{\cos }^{2}2x}}dx$

$dy=\frac{1}{2\sqrt{x}{\cos }^{2}x}dx$

$dy=\frac{1}{\sqrt{x}{\cos }^2\sqrt{x}}dx$

.$dy=\frac{1}{2\sqrt{x}\cos \sqrt{x}}dx$

$dy=\frac{1}{2\sqrt{x}{\cos }^2\sqrt{x}}dx$