Question

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

• A. $1-\frac{1}{x^2}$
• B. $1+\frac{1}{x^2}$
• C. $1-\frac{1}{2x}$
• D. none of these

Answer 1

The correct option is A

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

Explanation for correct option:

Step-1: Simplify the given data

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

$$\begin{gathered} =\frac{d}{dx}\left(x+\frac{1}{x}+2\sqrt{x}\times \frac{1}{\sqrt{x}}\right) \\ =\frac{d}{dx}\left(x+\frac{1}{x}+2\right) \end{gathered}$$

Step-2: Differentiate w.r.t. $x$.

$$\begin{gathered} =1+\left(-\frac{1}{x^2}\right)+0 \\ =1-\frac{1}{x^2} \end{gathered}$$

Hence, correct answer is option $\left(A\right)$