If \( y=\sqrt{x}+\displaystyle\frac{1}{\sqrt{x}} \) then \( 2x\displaystyle\frac{dy}{dx} \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{x}-\displaystyle\frac{1}{\sqrt{x}} \)
Writing \( y=x^{1/2}+x^{-1/2} \):
\( \displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}} \)
Multiplying by 2x:
\( \displaystyle 2x\frac{dy}{dx}=\frac{2x}{2\sqrt{x}}-\frac{2x}{2x\sqrt{x}} \)
\( \displaystyle =\sqrt{x}-\frac{1}{\sqrt{x}} \)
Why the other options are incorrect
Option (2) \( \sqrt{x}+\displaystyle\frac{1}{\sqrt{x}} \) — Incorrect. The derivative of \( x^{-1/2} \) is negative, so the second term must be subtracted.
Option (3) \( \displaystyle\frac{1}{\sqrt{x}}+\sqrt{x} \) — Incorrect. This is the same expression as Option (2), with a + sign where a − sign is needed.
Option (4) \( \displaystyle\frac{1}{2\sqrt{x}}+\displaystyle\frac{1}{2x\sqrt{x}} \) — Incorrect. This is \( \frac{dy}{dx} \) with a sign error, and not yet multiplied by 2x.