Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 34

Answer

$$f'\left( x \right) = \pi {x^{\pi - 1}}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^\pi } \cr & {\text{evaluate the derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{x^\pi }} \right) \cr & {\text{general power rule}} \cr & \frac{d}{{dx}}\left( {{x^p}} \right) = p{x^{p - 1}},{\text{ }}p = \pi \cr & then \cr & f'\left( x \right) = \pi {x^{\pi - 1}} \cr} $$
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