Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises: 41

Answer

\[{y^,} = \frac{1}{4} \cdot \,{\left( {\frac{{2x}}{{4x - 3}}} \right)^{ - \frac{3}{4}}} \cdot \frac{{ - 6}}{{\,{{\left( {4x - 3} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \sqrt[4]{{\frac{{2x}}{{4x - 3}}}} \hfill \\ \hfill \\ derivative \hfill \\ \hfill \\ {y^,} = \,{\left( {\frac{{2x}}{{4x - 3}}} \right)^{\frac{1}{4}}} \hfill \\ \hfill \\ use\,\,the\,\,chain\,rule \hfill \\ \hfill \\ {y^,} = \frac{1}{4} \cdot \,{\left( {\frac{{2x}}{{4x - 3}}} \right)^{ - \frac{3}{4}}} \cdot \,{\left( {\frac{{2x}}{{4x - 3}}} \right)^,} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = \frac{1}{4} \cdot \,{\left( {\frac{{2x}}{{4x - 3}}} \right)^{ - \frac{3}{4}}} \cdot \,\frac{{2\,\left( {4x - 3} \right) - 2x \cdot 4}}{{\,{{\left( {4x - 3} \right)}^2}}} \hfill \\ \hfill \\ {y^,} = \frac{1}{4} \cdot \,{\left( {\frac{{2x}}{{4x - 3}}} \right)^{ - \frac{3}{4}}} \cdot \frac{{ - 6}}{{\,{{\left( {4x - 3} \right)}^2}}} \hfill \\ \end{gathered} \]
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