Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 143: 88


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

Work Step by Step

$(a)$ $f(x)=x^n$. Differentiating with respect to $x$, $$f'(x)=nx^{n-1}$$Differentiating again, $$f''(x)=n(n-1)x^{n-2}$$Differentiating again, $$f'''(x)=n(n-1)(n-2)x^{n-3}$$From here we can see the pattern and suggest that, $$f^{(n)}(x)=n(n-1)(n-2)...x^{n-n}=n!x^0=n!$$ $(b)$ $f(x)=\frac1x=x^{-1}$. Differentiating with respect to $x$, $$f'(x)=-1\times x^{-2}=-x^{-2}$$ Differentiating again, $$f''(x)=(-1)(-2)x^{-3}=2x^{-3}$$Differentiating again, $$f'''(x)=(2)(-3)x^{-4}=-6x^{-4}$$From here we can observe the pattern, $$f^{(n)}=(-1)^nn!x^{-(n+1)}=(-1)^n\frac{n!}{x^{n+1}}$$
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