Tính vi phân của hàm số $y=\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}$

Chọn B

$\begin{array}{l}y\text{'}=\frac{1}{2\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}}.\left[\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)\right]\text{'}\\ =\frac{1}{2\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}}.c\text{os}\left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right).\left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)\text{'}\\ =\frac{c\text{os}\left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}{2\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}}.\frac{(\text{}c\text{osx - 2sin2x) .(cosx+sin 2x) - ( sin x +}c\text{os2x). (-sin}x+\text{}2\cos 2x)}{{(c\text{osx +sin2x)}}^2}\\ =\frac{c\text{os}\left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}{2\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}}.\frac{{\text{cos}}^{2}x-\sin 2x.c{\text{osx- 2sin}}^{2}2x-(-{\sin }^{2}x+\mathrm{s}\text{inx}.c{\text{os2x+2cos}}^{2}2x)}{{(c\text{osx +sin2x)}}^2}\\ =\frac{c\text{os}\left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}{2\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}}.\frac{-1-\sin 2x.c\text{osx-sinx}.c\text{os2x}}{{(c\text{osx +sin2x)}}^2}\end{array}$

Vi phân của hàm số là $dy=\frac{c\text{os}\left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}{2\sqrt{\sin \left(\frac{\sin x+\text{}c\text{os2x}}{\text{cosx + sin2x}}\right)}}.\frac{-1-\sin 2x.c\text{osx-sinx}.c\text{os2x}}{{(c\text{osx +sin2x)}}^2}dx$