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If $y=\frac{1}{x}+\cos 2 x$, then $\frac{d^2 y}{d x^2}$ is equal to
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$\frac{2}{x^3}+\frac{4}{x}-4 y$
$\begin{aligned} & y=\frac{1}{x}+\cos 2 x \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-1}{x^2}-2 \sin 2 x \Rightarrow \frac{d^2 y}{d x^2}=\frac{2}{x^3}-4 \cos 2 x \\ & =\frac{2}{x^3}-4\left(y-\frac{1}{x}\right)=\frac{2}{x^3}+\frac{4}{x}-4 y \\ & \end{aligned}$
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