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: 23

Answer

\[\frac{{dy}}{{dx}} = \frac{{5\sqrt {{x^4} + {y^2}} - 2{x^3}}}{{y - 6{y^2}\sqrt {{x^4} + {y^2}} }}\]

Work Step by Step

\[\begin{gathered} \sqrt {{x^4} + {y^2}} = 5x + 2{y^3} \hfill \\ \hfill \\ implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {\sqrt {{x^4} + {y^2}} } \right] = \frac{d}{{dx}}\,\,\left[ {5x + 2{y^3}} \right] \hfill \\ \hfill \\ then \hfill \\ \end{gathered} \] \[\frac{{4{x^3} + 2yy'}}{{2\sqrt {{x^4} + {y^2}} }} = 5 + 6{y^2}y'\] \[simplify\] \[\frac{{2{x^3} + yy'}}{{\sqrt {{x^4} + {y^2}} }} = 5 + 6{y^2}y'\] \[collect\,\,like\,\,terms\] \[\frac{{yy'}}{{\sqrt {{x^4} + {y^2}} }} - 6{y^2}y' = 5 - \frac{{2{x^3}}}{{\sqrt {{x^4} + {y^2}} }}\] \[factor\,\,{y^,}\] \[\left( {\frac{y}{{\sqrt {{x^4} + {y^2}} }} - 6{y^2}} \right)y' = 5 - \frac{{2{x^3}}}{{\sqrt {{x^4} + {y^2}} }}\] \[solve\,\,for\,\,{y^,}\] \[\frac{{dy}}{{dx}} = \frac{{5 - \frac{{2{x^3}}}{{\sqrt {{x^4} + {y^2}} }}}}{{\frac{y}{{\sqrt {{x^4} + {y^2}} }} - 6{y^2}}}\] \[\begin{gathered} simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{5\sqrt {{x^4} + {y^2}} - 2{x^3}}}{{y - 6{y^2}\sqrt {{x^4} + {y^2}} }} \hfill \\ \end{gathered} \]
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