y₁ = 1/x হলে, y_n = কত?

Explanation:

Another Explanation (5): y₁ = \(\frac{1}{x}\) হলে, y_n নির্ণয়: y₁ = x^-1 y₂ = \(\frac{dy_1}{dx}\) = -1 * x^-2 = \(\frac{-1}{x^2}\) y₃ = \(\frac{dy_2}{dx}\) = (-1)(-2) x^-3 = \(\frac{2}{x^3}\) y₄ = \(\frac{dy_3}{dx}\) = (2)(-3) x^-4 = \(\frac{-6}{x^4}\) .... y_n = \(\frac{d(y_{n-1})}{dx}\) লক্ষ্য করি, y₁ = \(\frac{1}{x}\) = \(\frac{(-1)^{1-1}(1-1)!}{x^1}\) = \(\frac{(-1)^0 0!}{x^1}\) = \(\frac{1}{x}\) 🥳 y₂ = \(\frac{-1}{x^2}\) = \(\frac{(-1)^{2-1}(2-1)!}{x^2}\) = \(\frac{(-1)^1 1!}{x^2}\) = \(\frac{-1}{x^2}\) y₃ = \(\frac{2}{x^3}\) = \(\frac{(-1)^{3-1}(3-1)!}{x^3}\) = \(\frac{(-1)^2 2!}{x^3}\) = \(\frac{2}{x^3}\) y₄ = \(\frac{-6}{x^4}\) = \(\frac{(-1)^{4-1}(4-1)!}{x^4}\) = \(\frac{(-1)^3 3!}{x^4}\) = \(\frac{-6}{x^4}\) সুতরাং, প্যাটার্ন অনুযায়ী, y_n = \(\frac{(-1)^{n-1} (n-1)!}{x^n}\) 🤔 অতএব, y_n = \(\frac{(-1)^{n-1}(n-1)!}{x^n}\) ✅