Finite Math and Applied Calculus (6th Edition)

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

Chapter 11 - Section 11.4 - The Chain Rule - Exercises - Page 831: 34

Answer

$$h'\left( x \right) = - 12.8{\left( {6.4x - 3} \right)^{ - 3}} - 8.6{\left( {4.3x - 1} \right)^{ - 3}}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {\left( {6.4x - 3} \right)^{ - 2}} + {\left( {4.3x - 1} \right)^{ - 2}} \cr & {\text{Differentiate}} \cr & h'\left( x \right) = \frac{d}{{dx}}\left( {{{\left( {6.4x - 3} \right)}^{ - 2}}} \right) + \frac{d}{{dx}}\left( {{{\left( {4.3x - 1} \right)}^{ - 2}}} \right) \cr & {\text{Use the chain rule and the power rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr & h'\left( x \right) = - 2{\left( {6.4x - 3} \right)^{ - 3}}\left( {6.4} \right) - 2{\left( {4.3x - 1} \right)^{ - 3}}\left( {4.3} \right) \cr & {\text{Simplifying}} \cr & h'\left( x \right) = - 12.8{\left( {6.4x - 3} \right)^{ - 3}} - 8.6{\left( {4.3x - 1} \right)^{ - 3}} \cr} $$
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