Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 35

Answer

$$ f'(t)= -\frac{ 3e^{-3t}}{(1-e^{-3t})^2}.$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(e^x)'=e^x$ Since we have $$ f(t)= \frac{1}{1-e^{-3t}}$$ then the derivative $ f'(t)$, using the chain and quotient rules, is given by $$ f'(t)= \frac{(1-e^{-3t})(1)'-(1-e^{-3t})'}{(1-e^{-3t})^2}=\frac{ -(0-e^{-3t}(-3))}{(1-e^{-3t})^2}\\ =-\frac{ 3e^{-3t}}{(1-e^{-3t})^2}.$$
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