# Chapter 3: Derivatives - Section 3.7 - Implicit Differentiation - Exercises 3.7 - Page 155: 28

$${m_1} = - 1{\text{ and }}{m_2} = 1$$

#### Work Step by Step

\eqalign{ & {\left( {{x^2} + {y^2}} \right)^2} = {\left( {x - y} \right)^2} \cr & {\text{then}} \cr & {x^2} + {y^2} = x - y \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}}\left( {x - y} \right) \cr & {\text{find derivatives}} \cr & 2x + 2y\frac{{dy}}{{dx}} = 1 - \frac{{dy}}{{dx}} \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & 2y\frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} = 1 - 2x \cr & \left( {2y + 1} \right)\frac{{dy}}{{dx}} = 1 - 2x \cr & \frac{{dy}}{{dx}} = \frac{{1 - 2x}}{{2y + 1}} \cr & \cr & {\text{Find the slope at the points }}\left( {1,0} \right){\text{ and }}\left( {1, - 1} \right) \cr & {\left. {m = \frac{{dy}}{{dx}}} \right|_{\left( {1,0} \right)}} = \frac{{1 - 2\left( 1 \right)}}{{2\left( 0 \right) + 1}} = - 1 \cr & {\left. {m = \frac{{dy}}{{dx}}} \right|_{\left( {1, - 1} \right)}} = \frac{{1 - 2\left( 1 \right)}}{{2\left( { - 1} \right) + 1}} = 1 \cr & \cr & {m_1} = - 1{\text{ and }}{m_2} = 1 \cr}

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