Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 150: 2

Answer

$f'(x) = cos (x) - x sin(x)+2sec^{2}(x)$

Work Step by Step

$f(x) = x cos(x) + 2tan (x)$ lets use the formulas $(fg)' = f'g+ fg'$, $(f+g)' = f'+ g'$, $(cf)' = cf'$ $f'(x) = (x cos(x) + 2tan (x))' = (x cos(x))' + (2tan (x))'$ $(2tan (x))'=2(tan (x))' =2sec^{2}(x)$ $(x cos(x))'= x' cos (x) + x (cos(x))' = cos (x) - x sin(x)$ So: $f'(x) = cos (x) - x sin(x)+2sec^{2}(x)$
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