Answer
$$54$$
Work Step by Step
Since $ y=4t^{-3}+3t^2$, then
$$\frac{d y}{d t}=-12 t^{-4}+6t, \quad \frac{d^2 y}{d t^2}=48 t^{-5}+6.$$
Hence, $
\left.\frac{d^{2} y}{d t^{2}}\right|_{t=1}=48+6=54
$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 135: 19
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