Chapter 6 - Applications of the Derivative - 6.4 Implicit Differentiation - 6.4 Exercises - Page 334: 23

$$y = \frac{{11}}{{12}}x - \frac{5}{6}$$

$$\eqalign{ & {e^{{x^2} + {y^2}}} = x{e^{5y}} - {y^2}{e^{5x/2}};{\text{ }}\left( {2,1} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{e^{{x^2} + {y^2}}}} \right) = \frac{d}{{dx}}\left( {x{e^{5y}} - {y^2}{e^{5x/2}}} \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( {{e^{{x^2} + {y^2}}}} \right) = \frac{d}{{dx}}\left( {x{e^{5y}}} \right) - \frac{d}{{dx}}\left( {{y^2}{e^{5x/2}}} \right) \cr & {\text{use product rule}} \cr & \frac{d}{{dx}}\left( {{e^{{x^2} + {y^2}}}} \right) = x\frac{d}{{dx}}\left( {{e^{5y}}} \right) + {e^{5y}}\frac{d}{{dx}}\left( x \right) - {e^{5x/2}}\frac{d}{{dx}}\left( {{y^2}} \right) - {y^2}\frac{d}{{dx}}\left( {{e^{5x/2}}} \right) \cr & {\text{use the chain rule}} \cr & {e^{{x^2} + {y^2}}}\frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = x{e^{5y}}\frac{d}{{dx}}\left( {5y} \right) + {e^{5y}}\left( 1 \right) - {e^{5x/2}}\left( {2y} \right)\frac{{dy}}{{dx}} - {y^2}\left( {\frac{5}{2}{e^{5x/2}}} \right) \cr & {\text{solve the derivatives}} \cr & {e^{{x^2} + {y^2}}}\left( {2x + 2y\frac{{dy}}{{dx}}} \right) = 5x{e^{5y}}\frac{{dy}}{{dx}} + {e^{5y}} - 2{e^{5x/2}}y\frac{{dy}}{{dx}} - \frac{5}{2}{y^2}{e^{5x/2}} \cr & 2x{e^{{x^2} + {y^2}}} + 2y{e^{{x^2} + {y^2}}}\frac{{dy}}{{dx}} = 5x{e^{5y}}\frac{{dy}}{{dx}} + {e^{5y}} - 2{e^{5x/2}}y\frac{{dy}}{{dx}} - \frac{5}{2}{y^2}{e^{5x/2}} \cr & 2y{e^{{x^2} + {y^2}}}\frac{{dy}}{{dx}} + 2{e^{5x/2}}y\frac{{dy}}{{dx}} - 5x{e^{5y}}\frac{{dy}}{{dx}} = {e^{5y}} - \frac{5}{2}{y^2}{e^{5x/2}} - 2x{e^{{x^2} + {y^2}}} \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & \left( {2y{e^{{x^2} + {y^2}}} + 2{e^{5x/2}}y - 5x{e^{5y}}} \right)\frac{{dy}}{{dx}} = {e^{5y}} - \frac{5}{2}{y^2}{e^{5x/2}} - 2x{e^{{x^2} + {y^2}}} \cr & {\text{multiply by 2}} \cr & \left( {4y{e^{{x^2} + {y^2}}} + 4{e^{5x/2}}y - 10x{e^{5y}}} \right)\frac{{dy}}{{dx}} = 2{e^{5y}} - 5{y^2}{e^{5x/2}} - 4x{e^{{x^2} + {y^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{2{e^{5y}} - 5{y^2}{e^{5x/2}} - 4x{e^{{x^2} + {y^2}}}}}{{4y{e^{{x^2} + {y^2}}} + 4{e^{5x/2}}y - 10x{e^{5y}}}} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {2,1} \right)}} = \frac{{2{e^{5\left( 1 \right)}} - 5{{\left( 1 \right)}^2}{e^{5\left( 2 \right)/2}} - 4\left( 2 \right){e^{{1^2} + {2^2}}}}}{{4\left( 1 \right){e^{{1^2} + {2^2}}} + 4{e^{5\left( 2 \right)/2}}y - 10\left( 2 \right){e^{5\left( 1 \right)}}}} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {2,1} \right)}} = \frac{{2{e^5} - 5{e^5} - 8{e^5}}}{{4{e^5} + 4{e^5} - 20{e^5}}} = \frac{{ - 11{e^5}}}{{ - 12{e^5}}} = \frac{{11}}{{12}} \cr & {\text{find the equation of the tangent line at the point }}\left( {2,1} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 1 = \frac{{11}}{{12}}\left( {x - 2} \right) \cr & y - 1 = \frac{{11}}{{12}}x - \frac{{11}}{6} \cr & y = \frac{{11}}{{12}}x - \frac{5}{6} \cr} $$