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: 36


$$ f'(t) = - (1-2t)e^{-2t}\sin(te^{-2t})$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(e^x)'=e^x$ Recall that $(\cos x)'=-\sin x$. Since we have $$ f(t)= \cos(te^{-2t})$$ then the derivative $ f'(t)$, using the chain and product rules, is given by $$ f'(t)= -\sin(te^{-2t}) (te^{-2t})'= - (e^{-2t}-2te^{-2t})\sin(te^{-2t})\\ = - (1-2t)e^{-2t}\sin(te^{-2t}).$$
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