#### 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) $

Published by
Pearson

ISBN 10:
0321947347

ISBN 13:
978-0-32194-734-5

$f'(x) = \pi(2^x+1)^{\pi-1} 2^x\ln(2) $

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