Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 192: 57

Answer

\[{y^,} = \frac{{5{x^4}}}{{\,{{\left( {x + 1} \right)}^6}}}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {\frac{x}{{x + 1}}} \right)^5} \hfill \\ \hfill \\ Differentiate\,\,both\,\,sides \hfill \\ \hfill \\ {y^,} = \frac{d}{{dx}}\,{\left( {\frac{x}{{x + 1}}} \right)^5} \hfill \\ \hfill \\ use\,\,the\,\,chain\,\,rule\,\,for\,\,powers \hfill \\ \hfill \\ \frac{d}{{dx}}\left( {g{{\left( x \right)}^n}} \right) = ng{\left( x \right)^{n - 1}}g'\left( x \right) \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ {y^,} = 5\,{\left( {\frac{x}{{x + 1}}} \right)^4} \cdot \,{\left( {\frac{x}{{x + 1}}} \right)^,} \hfill \\ \hfill \\ use\,\,the\,\,quotient\,\,rule \hfill \\ \hfill \\ {y^,} = 5\,{\left( {\frac{x}{{x + 1}}} \right)^4} \cdot \frac{{x + 1 - x}}{{\,{{\left( {x + 1} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = \frac{{5{x^4}}}{{\,{{\left( {x + 1} \right)}^6}}} \hfill \\ \end{gathered} \]
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