The value of k for which $k\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}},$ where $y=\sqrt{x}+\frac{1}{\sqrt{x}};$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $2x$
$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}$
so $2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}$
on comparison $k=2x$