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: 38

Answer

$$s'\left( x \right) = \frac{{90{{\left( {3x - 9} \right)}^2}}}{{{{\left( {2x + 4} \right)}^3}}}$$

Work Step by Step

$$\eqalign{ & s\left( x \right) = {\left( {\frac{{3x - 9}}{{2x + 4}}} \right)^3} \cr & {\text{Differentiate}} \cr & s'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {\frac{{3x - 9}}{{2x + 4}}} \right)}^3}} \right] \cr & {\text{Use the chain rule and the power rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr & s'\left( x \right) = 3{\left( {\frac{{3x - 9}}{{2x + 4}}} \right)^{3 - 1}}\frac{d}{{dx}}\left[ {\frac{{3x - 9}}{{2x + 4}}} \right] \cr & {\text{By the quotient rule}} \cr & s'\left( x \right) = 3{\left( {\frac{{3x - 9}}{{2x + 4}}} \right)^2}\left( {\frac{{\left( {2x + 4} \right)\left( 3 \right) - \left( {3x - 9} \right)\left( 2 \right)}}{{{{\left( {2x + 4} \right)}^2}}}} \right) \cr & s'\left( x \right) = 3{\left( {\frac{{3x - 9}}{{2x + 4}}} \right)^2}\left( {\frac{{6x + 12 - 6x + 18}}{{{{\left( {2x + 4} \right)}^2}}}} \right) \cr & s'\left( x \right) = 3{\left( {\frac{{3x - 9}}{{2x + 4}}} \right)^2}\left( {\frac{{30}}{{{{\left( {2x + 4} \right)}^2}}}} \right) \cr & s'\left( x \right) = \frac{{90{{\left( {3x - 9} \right)}^2}}}{{{{\left( {2x + 4} \right)}^3}}} \cr} $$
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