frac{d}{dx} {left( sqrt{x} + frac{1}{sqrt{x}} | vedclass

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Question

(C) Let $y = {\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)}^2$. Expanding the square using the identity ${(a+b)}^2 = a^2 + b^2 + 2ab$: $y = {(\sqrt{x}

Vedclass · Accepted Answer

$1 - \frac{1}{x^2}$

Vedclass · Answer

(C) Let $y = {\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)}^2$. Expanding the square using the identity ${(a+b)}^2 = a^2 + b^2 + 2ab$: $y = {(\sqrt{x})}^2 + {\left( \frac{1}{\sqrt{x}} \right)}^2 + 2(\sqrt{x})\left( \frac{1}{\sqrt{x}} \right)$ $y = x + \frac{1}{x} + 2$. Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(2)$ $\frac{dy}{dx} = 1 + (-1)x^{-2} + 0$ $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.

$\frac{d}{dx} {\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)}^2$ is equal to

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