If y=1x+cos2x, then d2ydx2=

y=1x+cos 2xdxdy=-1x2-2sin 2xd2xdx2=2x3-4cos 2x =2x3-4(y-1x) =2x3-4y+4x

If y=1x+cos2x, then d2ydx2=

y=1x+cos 2xdxdy=-1x2-2sin 2xd2xdx2=2x3-4cos 2x =2x3-4(y-1x) =2x3-4y+4x

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If $y = \frac{1}{x} + \cos 2 x, t h e n \frac{d^{2} y}{d x^{2}} =$

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$\frac{2}{x^{3}} + \frac{4}{x} - 4 y$

$4 y - \frac{4}{x} - \frac{2}{x^{3}}$

$4 y - \frac{4}{x} - \frac{1}{x^{3}}$

$\frac{2}{x^{3}} + 4 y - \frac{4}{x}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} y & = & \frac{1}{x} + \cos 2 x \\ \frac{dx}{dy} & = & \frac{- 1}{x^{2}} - 2 \sin 2 x \\ \frac{d^{2} x}{{dx}^{2}} & = & \frac{2}{x^{3}} - 4 \cos 2 x \\ = & \frac{2}{x^{3}} - 4 (y - \frac{1}{x}) \\ = & \frac{2}{x^{3}} - 4 y + \frac{4}{x} \end{array}$