MCQ
If $y =\sqrt{x+\frac{1}{x}}$, then $\frac{d y}{d x}=$ ?
  • A
    $\frac{x^2-1}{2 x^2 \sqrt{x^2+1}}$
  • B
    $\frac{1-x^2}{2 x^2 \sqrt{x^2+1}}$
  • $\frac{x^2-1}{2 x \sqrt{x} \sqrt{x^2+1}}$
  • D
    $\frac{1-x^2}{2 x \sqrt{x} \sqrt{x^2+1}}$

Answer

Correct option: C.
$\frac{x^2-1}{2 x \sqrt{x} \sqrt{x^2+1}}$
(C) $\frac{x^2-1}{2 x \sqrt{x} \sqrt{x^2+1}}$
Hint:
$
\begin{gathered}
\frac{d y}{d x}=\frac{1}{2 \sqrt{x+\frac{1}{x}}} \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right) \\
=\frac{\sqrt{x}}{2 \sqrt{x^2+1}}\left(1-\frac{1}{x^2}\right)=\frac{x^2-1}{2 x \sqrt{x} \sqrt{x^2+1}}
\end{gathered}
$

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