y=ln{e^x((x-1)/(x+1))^(5/2)} হলে, (dy)/(dx) = কত?

Explanation:

Another Explanation (5): bài toán: \(y=\ln\left\{e^x\left(\frac{x-1}{x+1}\right)^{\frac{5}{2}}\right\}\) হলে, \(\frac{dy}{dx} = কত?🤔 সমাধান: প্রথমে, লগারিদমের সূত্র ব্যবহার করে \(y\) কে সরল করি: \(y = \ln(e^x) + \ln\left(\left(\frac{x-1}{x+1}\right)^{\frac{5}{2}}\right)\) \(y = x + \frac{5}{2} \ln\left(\frac{x-1}{x+1}\right)\) \(y = x + \frac{5}{2} \left[\ln(x-1) - \ln(x+1)\right]\) এখন, \(x\) এর সাপেক্ষে অন্তরীকরণ করি: \(\frac{dy}{dx} = \frac{d}{dx}\left[x + \frac{5}{2} \left(\ln(x-1) - \ln(x+1)\right)\right]\) \(\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{5}{2} \left[\frac{d}{dx}\ln(x-1) - \frac{d}{dx}\ln(x+1)\right]\) \(\frac{dy}{dx} = 1 + \frac{5}{2} \left[\frac{1}{x-1} - \frac{1}{x+1}\right]\) \(\frac{dy}{dx} = 1 + \frac{5}{2} \left[\frac{(x+1) - (x-1)}{(x-1)(x+1)}\right]\) \(\frac{dy}{dx} = 1 + \frac{5}{2} \left[\frac{x+1 - x+1}{x^2-1}\right]\) \(\frac{dy}{dx} = 1 + \frac{5}{2} \left[\frac{2}{x^2-1}\right]\) \(\frac{dy}{dx} = 1 + \frac{5}{x^2-1}\) \(\frac{dy}{dx} = \frac{x^2-1+5}{x^2-1}\) \(\frac{dy}{dx} = \frac{x^2+4}{x^2-1}\) ✅ অতএব, \(\frac{dy}{dx} = \frac{x^2+4}{x^2-1}\) 😎