যদি  x^py^q=(x+y)^(p+q)  হয় তাহলে  (dy)/(dx)=?   

Explanation:

Another Explanation (5): যদি \(x^p y^q = (x+y)^{p+q}\) হয়, তবে \(\frac{dy}{dx}\) নির্ণয় করতে হবে। উভয় পক্ষে \(ln\) নিয়ে পাই, \(ln(x^p y^q) = ln((x+y)^{p+q})\) \(ln(x^p) + ln(y^q) = (p+q)ln(x+y)\) \(p \cdot ln(x) + q \cdot ln(y) = (p+q)ln(x+y)\) এখন, \(x\) এর সাপেক্ষে অন্তরীকরণ করে পাই, \(\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = (p+q) \cdot \frac{1}{x+y} \cdot (1 + \frac{dy}{dx})\) \(\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = \frac{p+q}{x+y} + \frac{p+q}{x+y} \frac{dy}{dx}\) \(\frac{dy}{dx} \left( \frac{q}{y} - \frac{p+q}{x+y} \right) = \frac{p+q}{x+y} - \frac{p}{x}\) \(\frac{dy}{dx} \left( \frac{q(x+y) - (p+q)y}{y(x+y)} \right) = \frac{(p+q)x - p(x+y)}{x(x+y)}\) \(\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x+y)} \right) = \frac{px + qx - px - py}{x(x+y)}\) \(\frac{dy}{dx} \left( \frac{qx - py}{y(x+y)} \right) = \frac{qx - py}{x(x+y)}\) \(\frac{dy}{dx} = \frac{qx - py}{x(x+y)} \cdot \frac{y(x+y)}{qx - py}\) \(\frac{dy}{dx} = \frac{y}{x}\) 🎉