Question

If $y=\sqrt{x}-\frac{1}{\sqrt{x}}$, then $2x\frac{dy}{dx}+y$=

• A. $\sqrt{x}$
• B. $2\sqrt{x}$
• C. $3\sqrt{x}$
• D. $\frac{\sqrt{x}}{2}$

Answer 1

Answer: (B)

$2\sqrt{x}$

$y=\sqrt{x}-\frac{1}{\sqrt{x}}\frac{dy}{dx}=\frac{d}{dx}(x{)}^{\frac{1}{2}}-\frac{d}{dx}(\frac{1}{\sqrt{x}})=\frac{1}{2\sqrt{x}}-(-\frac{1}{2})x^{-\frac{3}{2}}=\frac{1}{2\sqrt{x}}+\frac{x^{-\frac{3}{2}}}{2}\frac{2xdy}{dx}+y=2x[\frac{1}{2\sqrt{x}}+\frac{x^{-\frac{3}{2}}}{2}]+\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x}-\frac{1}{\sqrt{x}}=2\sqrt{x}$