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

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Question

(A) Let $y = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2$. Expanding the expression using $(a+b)^2 = a^2 + 2ab + b^2$,we get: $y = (\sqrt{x})^2 + 2(\sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $y = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2$. Expanding the expression using $(a+b)^2 = a^2 + 2ab + b^2$,we get: $y = (\sqrt{x})^2 + 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2$ $y = x + 2 + \frac{1}{x}$ Now,differentiate with respect to $x$: $\frac{d y}{d x} = \frac{d}{d x}(x) + \frac{d}{d x}(2) + \frac{d}{d x}(x^{-1})$ $\frac{d y}{d x} = 1 + 0 - x^{-2}$ $\frac{d y}{d x} = 1 - \frac{1}{x^2}$ Thus,the correct option is $A$.

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

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