xy+x^2y^2-c=0  হলে 

dx/dy=?

দেওয়া আছে, \(xy + x^2y^2 - c = 0\)

আমরা \(y\) এর সাপেক্ষে উভয় দিকে অন্তরকলন করব।

সুতরাং, \(\frac{d}{dy}(xy + x^2y^2 - c) = \frac{d}{dy}(0)\)

\(\implies \frac{d}{dy}(xy) + \frac{d}{dy}(x^2y^2) - \frac{d}{dy}(c) = 0\)

এখন, \(\frac{d}{dy}(xy) = x\frac{dy}{dy} + y\frac{dx}{dy} = x + y\frac{dx}{dy}\) 🤓

এবং, \(\frac{d}{dy}(x^2y^2) = x^2\frac{d}{dy}(y^2) + y^2\frac{d}{dy}(x^2) = x^2(2y) + y^2(2x\frac{dx}{dy}) = 2x^2y + 2xy^2\frac{dx}{dy}\) ✨

যেহেতু \(c\) একটি ধ্রুবক, তাই \(\frac{d}{dy}(c) = 0\) 🎉

সুতরাং, \(x + y\frac{dx}{dy} + 2x^2y + 2xy^2\frac{dx}{dy} = 0\)

\(\implies \frac{dx}{dy}(y + 2xy^2) = -x - 2x^2y\)

\(\implies \frac{dx}{dy} = \frac{-x - 2x^2y}{y + 2xy^2}\)

\(\implies \frac{dx}{dy} = \frac{-x(1 + 2xy)}{y(1 + 2xy)}\)

\(\implies \frac{dx}{dy} = -\frac{x}{y}\) 🥳