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