If y = frac{1}{x} + cos 2x,then frac{d^2y}{dx | vedclass

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Question

(C) Given $y = \frac{1}{x} + \cos 2x$. First,differentiate with respect to $x$: $\frac{dy}{dx} = -\frac{1}{x^2} - 2 \sin 2x$. Now,differentiate again

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $y = \frac{1}{x} + \cos 2x$. First,differentiate with respect to $x$: $\frac{dy}{dx} = -\frac{1}{x^2} - 2 \sin 2x$. Now,differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{2}{x^3} - 4 \cos 2x$. From the original equation,we have $\cos 2x = y - \frac{1}{x}$. Substituting this into the second derivative: $\frac{d^2y}{dx^2} = \frac{2}{x^3} - 4(y - \frac{1}{x}) = \frac{2}{x^3} - 4y + \frac{4}{x}$. Rearranging the terms,we get $\frac{2}{x^3} + \frac{4}{x} - 4y$.

If $y = \frac{1}{x} + \cos 2x$,then $\frac{d^2y}{dx^2}$ is equal to

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