Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 200: 24

$y' = - \frac{1}{{2y - 2\cos y\sqrt {x + {y^2}} }}$

Work Step by Step

$find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.$ $\frac{1}{{2\sqrt {x + {y^2}} }}\, \cdot \,\,\,\,{\left( {x + {y^2}} \right)^\prime } = \cos y \cdot y'$ $then$ $\frac{1}{{2\sqrt {x + {y^2}} \, \cdot \,\left( {1 + 2yy'} \right)}} = y'\cos y$ $simplify$ $\frac{{2yy'}}{{2\sqrt {x + {y^2}} }} - y'\cos y = - \frac{1}{{2\sqrt {x + {y^2}} }}$ $factor\,\,y'$ $y'\,\left( {\frac{y}{{\sqrt {x + {y^2}} }} - \cos y} \right) = - \frac{1}{{2\sqrt {x + {y^2}} }}$ $solve\,\,for\,\,y'$ $\begin{gathered} y' = - \frac{{\frac{1}{{2\sqrt {x + {y^2}} }}}}{{\frac{y}{{\sqrt {x + {y^2}} }} - \cos y}} \hfill \\ \hfill \\ y' = - \frac{1}{{2y - 2\cos y\sqrt {x + {y^2}} }} \hfill \\ \end{gathered}$

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