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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given expression: $y = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2$ 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 $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(2)$ $\frac{dy}{dx} = 1 - x^{-2} + 0$ $\frac{dy}{dx} = 1 - \frac{1}{x^2}$ Thus,the correct option is $A$.

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

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