Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 11 - Section 11.1 - Derivatives of Powers, Sums, and Constant Multiples - Exercises - Page 795: 49

Answer

$\displaystyle \frac{ds}{dt}=-2.31x^{-2.1}-1.2x^{-0.4}$

Work Step by Step

SUMMARY (rules in differential notation): 1. The Power Rule$:\ \ \ \displaystyle \frac{d}{dx}[x^{n}]=n\cdot x^{n-1 } $ 2. Sum Rule: $\displaystyle \ \ \ \frac{d}{dx}[f\pm g](x)=\frac{d}{dx}[f(x)]\pm\frac{d}{dx}[g(x)] $ 3. Constant Multiple Rule:$\ \ \displaystyle \frac{d}{dx}[cf(x)]=c\cdot\frac{d}{dx}[f(x)] $ 4. Constant times x:$\ \ \ \displaystyle \frac{d}{dx}(cx)=c $ 5. Constant:$\displaystyle \ \ \ \ \ \frac{d}{dx}(c)=0 $ ------------------ $ \displaystyle \frac{ds}{dt}= \frac{d}{dx}[ 2.3+2.1t^{-1.1}-2t^{0.6})$ = $\ \ \ $...(2) $=\displaystyle \frac{d}{dt}(2.3)+\frac{d}{dt}(2.1t^{-1.1})-\frac{d}{dt}(2t^{0.6})$ = $\ \ \ $...($3$) $=2.3\displaystyle \frac{d}{dx}( x^{0})+2.1\frac{d}{dx}(x^{-1.1})-2\frac{d}{dt}(t^{0.6})$ = $\ \ \ $...($1$) $=0+2.1(-1.1x^{-2.1})-2(0.6x^{-0.4})$ $\displaystyle \frac{ds}{dt}=-2.31x^{-2.1}-1.2x^{-0.4}$
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