y= 1/x  হলে y₉₈=? 

y = \(\frac{1}{x}\)

y = \(x^{-1}\)

প্রথম অন্তরকলজ:

y₁ = \(\frac{dy}{dx}\) = \(-1 \cdot x^{-2}\) = \(\frac{-1}{x^2}\) = \(\frac{(-1)^1 \cdot 1!}{x^{1+1}}\)

দ্বিতীয় অন্তরকলজ:

y₂ = \(\frac{d^2y}{dx^2}\) = \((-1)(-2) \cdot x^{-3}\) = \(\frac{2}{x^3}\) = \(\frac{(-1)^2 \cdot 2!}{x^{2+1}}\)

তৃতীয় অন্তরকলজ:

y₃ = \(\frac{d^3y}{dx^3}\) = \(2(-3) \cdot x^{-4}\) = \(\frac{-6}{x^4}\) = \(\frac{(-1)^3 \cdot 3!}{x^{3+1}}\)

অতএব, n-তম অন্তরকলজ:

y_n = \(\frac{d^ny}{dx^n}\) = \(\frac{(-1)^n \cdot n!}{x^{n+1}}\)

সুতরাং, 98-তম অন্তরকলজ:

y₉₈ = \(\frac{d^{98}y}{dx^{98}}\) = \(\frac{(-1)^{98} \cdot 98!}{x^{98+1}}\) = \(\frac{98!}{x^{99}}\) 🎉