d/dx(sqrtx)^sqrtx = কত?

Explanation:

Another Explanation (5): \( \frac{d}{dx} (\sqrt{x})^{\sqrt{x}} \) নির্ণয়: ধরি, \( y = (\sqrt{x})^{\sqrt{x}} \) উভয়পক্ষে স্বাভাবিক লগারিদম নিয়ে পাই, \( \ln y = \ln (\sqrt{x})^{\sqrt{x}} \) \( \ln y = \sqrt{x} \ln (\sqrt{x}) \) \( \ln y = \sqrt{x} \ln (x^{\frac{1}{2}}) \) \( \ln y = \frac{1}{2} \sqrt{x} \ln x \) এখন, x এর সাপেক্ষে অন্তরীকরণ করে পাই, \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \sqrt{x} \cdot \frac{1}{x} + \ln x \cdot \frac{1}{2\sqrt{x}} \right] \) \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}} \right] \) \( \frac{dy}{dx} = \frac{y}{2} \left[ \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}} \right] \) \( \frac{dy}{dx} = \frac{(\sqrt{x})^{\sqrt{x}}}{2} \left[ \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}} \right] \) \( \frac{dy}{dx} = \frac{(\sqrt{x})^{\sqrt{x}}}{2\sqrt{x}} \left[ 1 + \frac{\ln x}{2} \right] \) \( \frac{dy}{dx} = \frac{(\sqrt{x})^{\sqrt{x}}}{\sqrt{x}} \cdot \frac{1}{2} \left[ 1 + \ln \sqrt{x} \right] \) \( \frac{dy}{dx} = (\sqrt{x})^{\sqrt{x}-1} \cdot \frac{1}{2} \left[ 1 + \ln \sqrt{x} \right] \) \( \frac{dy}{dx} = \frac{1}{2} (\sqrt{x})^{\sqrt{x}-1} \left[ 1 + \ln \sqrt{x} \right] \) সুতরাং, \( \frac{d}{dx} (\sqrt{x})^{\sqrt{x}} = \frac{1}{2} (\sqrt{x})^{\sqrt{x}-1} (1 + \ln \sqrt{x}) \) 🥳