Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.3 Introduction To Techniques Of Differentiation - Exercises Set 2.3 - Page 140: 26

Answer

An approximation for $f′(1)=-1.997$ The actual value of $f′(1)=-2$ The two values are comparable. Therefore, we confirm both the approximation and the actual value of f′(1)

Work Step by Step

We approximate $f'(1)$ using the difference quotient $\frac{f(1+h)-f(1)}{h}$ for values of $h$ near $0$. In this case, we choose $h=0.001$. Approximate value of $f'(1) = \frac{f(1+h)-f(1)}{h}=\frac{f(1+0.001)-f(1)}{0.001} = \frac{\frac{1}{(1.001)^2}-\frac{1}{1^2}}{0.001} = -1.997$ We apply the power rule to take the derivative and find the real value $f'(1)$: Real value of $f'(1) = -2x^{-3} = -2(1)^{-3}= -2$
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