If y = x + frac{1}{x},then | vedclass

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(C) Given $y = x + \frac{1}{x}$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$$\frac{dy}{dx}

Vedclass · Accepted Answer

$x^2 \frac{dy}{dx} - xy + 2 = 0$

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(C) Given $y = x + \frac{1}{x}$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$$\frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^2}$.Now,multiply both sides by $x^2$:$x^2 \frac{dy}{dx} = x^2(1 - \frac{1}{x^2}) = x^2 - 1$.We know $y = x + \frac{1}{x}$,so $xy = x(x + \frac{1}{x}) = x^2 + 1$.From this,$x^2 = xy - 1$.Substituting $x^2$ into the equation $x^2 \frac{dy}{dx} = x^2 - 1$:$x^2 \frac{dy}{dx} = (xy - 1) - 1$$x^2 \frac{dy}{dx} = xy - 2$$x^2 \frac{dy}{dx} - xy + 2 = 0$.

If $y = x + \frac{1}{x}$,then

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