Answer
$0.035$
Work Step by Step
$f(x)$ = $tanx$
$f'(x)$ = $sec^{2}x$
$f'(x)$ = $\frac{f(x+h)-f(x)}{h}$
$x$ = $0$, $h$ = $2$
$sec^{2}0°$ = $\frac{tan2°-tan0°}{2°}$
$sec^{2}0°$ = $\frac{tan2°-tan0°}{2(\frac{\pi}{180})}$
$tan2°$ = $0.035$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 2 - Derivatives - 2.9 Linear Approximations and Differentials - 2.9 Exercises - Page 193: 27
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