y=(√x+1/√x) and (2xdy/dx+y)=? 

Another Explanation (5): প্রশ্ন: \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \) এবং \( 2x \frac{dy}{dx} + y = ? \) উত্তর: \( 2 \sqrt{x} \) সমাধান: প্রথমে \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \) কে লিখি: \[ y = x^{\frac{1}{2}} + x^{-\frac{1}{2}} \] ডিফারেনশিয়েশন: \[ \frac{dy}{dx} = \frac{1}{2} x^{-\frac{1}{2}} - \frac{1}{2} x^{-\frac{3}{2}} \] \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{x}} - \frac{1}{2} \cdot \frac{1}{x \sqrt{x}} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} \] দুটি ভগ্নাংশের সাধারণ সূত্র: \[ \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} \] \[ = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} \] সাধারণ মান: \[ \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} \] উপাদান হিসেবে: \[ \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} \] এখন, আমরা \( 2x \frac{dy}{dx} + y \) হিসাব করবো: \[ 2x \frac{dy}{dx} + y \] প্রথমে \( 2x \frac{dy}{dx} \): \[ 2x \left( \frac{1}{2 \sqrt{x}} - \frac{1}{2 x \sqrt{x}} \right) = 2x \cdot \frac{1}{2 \sqrt{x}} - 2x \cdot \frac{1}{2 x \sqrt{x}} \] সরলীকরণ: \[ = x \cdot \frac{1}{\sqrt{x}} - x \cdot \frac{1}{x \sqrt{x}} = \frac{x}{\sqrt{x}} - \frac{x}{x \sqrt{x}} \] \[ = \sqrt{x} - \frac{1}{\sqrt{x}} \] এখন, যোগ করি \( y \): \[ \sqrt{x} - \frac{1}{\sqrt{x}} + \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) = \sqrt{x} + \sqrt{x} - \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}} \] সাধারণীকরণ: \[ = 2 \sqrt{x} + 0 = 2 \sqrt{x} \] অতএব, \[ \boxed{ 2x \frac{dy}{dx} + y = 2 \sqrt{x} } \]