xy + x²y²  - c = 0  হলে  (dx)/(dy)  =?  

Explanation:

Another Explanation (5): দেওয়া আছে, \(xy + x^2y^2 - c = 0\) \(x\) এর সাপেক্ষে উভয় দিকে অন্তরকলন করে পাই, \(\frac{d}{dy}(xy + x^2y^2 - c) = \frac{d}{dy}(0)\) \(\implies \frac{d}{dy}(xy) + \frac{d}{dy}(x^2y^2) - \frac{d}{dy}(c) = 0\) \(\implies \left(x \cdot 1 + y \cdot \frac{dx}{dy}\right) + \left(x^2 \cdot 2y + y^2 \cdot 2x \frac{dx}{dy}\right) - 0 = 0\) \(\implies x + y\frac{dx}{dy} + 2x^2y + 2xy^2\frac{dx}{dy} = 0\) \(\implies \frac{dx}{dy}(y + 2xy^2) = -x - 2x^2y\) \(\implies \frac{dx}{dy} = \frac{-x - 2x^2y}{y + 2xy^2}\) \(\implies \frac{dx}{dy} = \frac{-x(1 + 2xy)}{y(1 + 2xy)}\) \(\implies \frac{dx}{dy} = -\frac{x}{y}\) সুতরাং, \(\frac{dx}{dy} = -\frac{x}{y}\) 🥳