Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.1 Exercises: 22

Answer

$f′(x)=\frac{-2}{x^{3}}$

Work Step by Step

$f(x)=7$ $f′(x)=\lim\limits_{h \to 0}\frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}$ $f′(x)=\lim\limits_{h \to 0}\frac{x^{2}-(x+h)^{2}}{(x)^{2}(x+h)^{2}(h)}$ $f′(x)=\lim\limits_{h \to 0}\frac{x^{2}-x^{2}-2xh+h^{2}}{(x)^{2}(x+h)^{2}(h)}$ $f′(x)=\lim\limits_{h \to 0}\frac{h(-2x+h)}{(x)^{2}(x+h)^{2}(h)}$ $f′(x)=\lim\limits_{h \to 0}\frac{-2x+h}{(x)^{2}(x+h)^{2}}$ $f′(x)=\frac{-2x+0}{x^{2}(x+0)^{2}}$ $f′(x)=\frac{-2}{x^{3}}$
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