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: 50

Answer

\[y = - \frac{2}{9}x + \frac{{17}}{9}\]

Work Step by Step

\[\begin{gathered} x{y^{\frac{5}{2}}} + {x^{\frac{3}{2}}}y = 12\,\,\,\,,\,\,\,\left( {4,1} \right) \hfill \\ \hfill \\ use\,\,the\,\,Implicit\,\,differentiation \hfill \\ \hfill \\ x\,\left( {\frac{5}{2}} \right){y^{\frac{3}{2}}}{y^,} + {y^{\frac{5}{2}}} + {x^{\frac{3}{2}}}{y^,} + \frac{3}{2}{x^{\frac{1}{2}}}y = 0 \hfill \\ \hfill \\ Collect\,\,like\,\,terms \hfill \\ \hfill \\ {y^,}\,\left( {\frac{5}{2}x{y^{\frac{3}{2}}} + {x^{\frac{3}{2}}}} \right) = - {y^{\frac{5}{2}}} - \frac{3}{2}{x^{\frac{1}{2}}}y \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {y^,} = \frac{{ - {y^{\frac{5}{2}}} - \frac{3}{2}{x^{\frac{1}{2}}}y}}{{\frac{5}{2}x{y^{\frac{3}{2}}} + {x^{\frac{3}{2}}}}} \hfill \\ \hfill \\ evaluate\,\,\left( {4,1} \right) \hfill \\ \hfill \\ {y^,} = \frac{{\, - {{\left( 1 \right)}^{\frac{1}{2}}} - \frac{3}{2}\,{{\left( 4 \right)}^{\frac{1}{2}}}\,\left( 1 \right)}}{{\frac{5}{2}\,\left( 4 \right)\,{{\left( 1 \right)}^{\frac{3}{2}}} + \,{{\left( 4 \right)}^{\frac{3}{2}}}}} \hfill \\ \hfill \\ use\,\,the\,\,point\,\,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ y - 1 = - \frac{2}{9}\,\left( {x - 4} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ y - 1 = - \frac{2}{9}x + \frac{8}{9} \hfill \\ \hfill \\ y = - \frac{2}{9}x + \frac{{17}}{9} \hfill \\ \end{gathered} \]
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