Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 60

Answer

$f^{1000}(x)=-1000e^{-x}+xe^{-x}$

Work Step by Step

$f(x)=xe^{-x}$ $f'(x)=\frac{d}{dx}xe^{-x}$ $f'(x)=e^{-x}\frac{d}{dx}(x)+x\frac{d}{dx}(e^{-x})$ $f'(x)=e^{-x}-x(e^{-x})$ Again differentiate with rest to x, we get $f''(x)=-2e^{-x}+x(e^{-x})$ Take third derivative of the function. $f'''(x)=3e^{-x}-x(e^{-x})$ Proceeding this process in the same way up to nth derivative, we observe the expression for nth derivative can be calculated as follows: $f^{n}(x)=(-1)^{(n-1)}(ne^{-x})+(-1)^{n}(xe^{-x})$ Take n = 1000, therefore, $f^{1000}(x)=(-1)^{(1000-1)}(1000e^{-x})+(-1)^{1000}(xe^{-x})$ $f^{1000}(x)=(-1)^{(999)}(1000e^{-x})+(-1)^{1000}(xe^{-x})$ Hence, $f^{1000}(x)=-1000e^{-x}+xe^{-x}$
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