Finite Math and Applied Calculus (6th Edition)

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

Chapter 11 - Section 11.3 - The Product and Quotient Rules - Exercises - Page 818: 61

Answer

3

Work Step by Step

(the derivative evaluated at $t=1$) Use the product rule, $\displaystyle \frac{d}{dt}[f(t)g(t)]=f^{\prime}(t)g(t)+f(t)g^{\prime}(t)$. $f(t)=t^{2}-t^{0.5},\quad f^{\prime}(t)=2t-0.5t^{-0.5}$ $g(t)=t^{0.5}+t^{-0.5},\quad g^{\prime}(t)=0.5t^{-0.5}-0.5t^{-1.5}$ $\displaystyle \frac{dy}{dt}=(2t-0.5t^{-0.5})(t^{0.5}+t^{-0.5})+(t^{2}-t^{0.5})(0.5t^{-0.5}-0.5t^{-1.5})$ $\left.\frac{dy}{dt}\right|_{t=1}=(2-0.5)(1+1)+(1-1)(0.5-0.5)$ $=1.5\cdot 2+0=3$
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