Answer
$f'(x) = \pi(2^x+1)^{\pi-1} 2^x\ln(2) $
Work Step by Step
If $f(x) = b^x$, then $f'(x) = b^x\ln(b)$
$f(x) = (2^x+1)^\pi$
Using General Power Rule and Chain Rule:
$f'(x) = \pi(2^x+1)^{\pi-1} 2^x\ln(2) $
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 211: 44
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