Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 66

Answer

(a)$$f^n(x)=n!$$ (b)$$f^n(x)=\frac{(-1)^nn!}{x^{n+1}}$$

Work Step by Step

(a) $$f(x)=x^n$$ - First derivative : $f'(x)=\frac{d}{dx}(x^n)=nx^{n-1}$ - Second derivative : $f''(x)=n\frac{d}{dx}(x^{n-1})=n(n-1)x^{n-2}$ - Third derivative : $f'''(x)=n(n-1)\frac{d}{dx}(x^{n-2})=n(n-1)(n-2)x^{n-3}$ Observing the pattern, we predict the nth derivative would be $$f^n(x)=n\times(n-1)\times...\times[n-(n-3)]\times[n-(n-2)]\times[n-(n-1)]x^{n-n}$$ $$f^n(x)=n\times(n-1)\times...\times3\times2\times1\times x^0$$ $$f^n(x)=n\times(n-1)\times...\times3\times2\times1$$ $$f^n(x)=n!$$ (b) $$f(x)=\frac{1}{x}=x^{-1}$$ - First derivative: $f'(x)=\frac{d}{dx}(x^{-1})=-x^{-2}$ - Second derivative: $f''(x)=-\frac{d}{dx}(x^{-2})=-(-2)x^{-3}=2x^{-3}$ - Third derivative: $f'''(x)=2\frac{d}{dx}x^{-3}=2\times(-3)x^{-4}$ Observing the pattern, we predict the nth derivative would be $$f^n(x)=(-1)\times(-2)\times(-3)\times...\times(-n)x^{-(n+1)}$$ $$f^n(x)=\frac{(-1)\times(-2)\times(-3)\times...\times(-n)}{x^{n+1}}$$ $$f^n(x)=\frac{(-1)^n(1\times2\times3\times...\times n)}{x^{n+1}}$$ $$f^n(x)=\frac{(-1)^nn!}{x^{n+1}}$$
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