Explanation Given information: f ( t ) = 2 t 3 − 4 t 2 + 3 t − 1 Formula used: Formulas of derivatives are as follows: d d x ( c ) = 0 , where c is a constant. d d x ( x n ) = n x n − 1 d d x { c . f ( x ) } = c . d d x f ( x ) Calculation: f ( t ) = 2 t 3 − 4 t 2 + 3 t − 1 (Given) Differentiating f ( t ) with respect to ‘ t’, we get f ' ( t ) = d d t f ( t ) = d d t ( 2 t 3 − 4 t 2 + 3 t − 1 ) = d d t ( 2 t 3 ) − d d t ( 4 t 2 ) + d d t ( 3 t ) − d d t ( 1 ) = 2 d d t ( t 3 ) − 4 d d t ( t 2 ) + 3 d d t ( t ) − d d t ( 1 ) = 2 ( 3 t 3 − 1 ) − 4 ( 2 t 2 − 1 ) + 3 ( 1 t 1 − 1 ) − 0 ( d

6th Edition

ISBN: 9781119399322

Author: Hughes-Hallett

Publisher: WILEY

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Chapter 3.1, Problem 51P

To determine

To find:

f'(t) and f''(t) of f(t)=2t3−4t2+3t−1.

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Chapter 3.1, Problem 50P

Chapter 3.1, Problem 52P

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Chapter 3 Solutions

APPLIED CALCULUS (WILEY PLUS)

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