Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 202: 85

$$y'' = \frac{{2{y^2}\left( {5 + 8x\sqrt y } \right)}}{{{{\left( {1 + 2x\sqrt y } \right)}^3}}}$$

$$\eqalign{ & \sqrt y + xy = 1 \cr & {y^{1/2}} + xy = 1 \cr & {\text{Differentiate both sides with respect to }}x \cr & \frac{1}{2}{y^{ - 1/2}}y' + xy' + y = 0 \cr & {\text{Solve for }}y' \cr & \frac{1}{2}{y^{ - 1/2}}y' + xy' = - y \cr & \left( {\frac{1}{2}{y^{ - 1/2}} + x} \right)y' = - y \cr & y' = - \frac{y}{{\frac{1}{2}{y^{ - 1/2}} + x}} \cr & {\text{Multiply numerator and denominator by 2}}{y^{1/2}} \cr & y' = - \frac{{2{y^{3/2}}}}{{1 + 2x{y^{1/2}}}} \cr & {\text{Differentiate both sides with respect to }}x \cr & y'' = - \frac{d}{{dx}}\left[ {\frac{{2{y^{3/2}}}}{{1 + 2x{y^{1/2}}}}} \right] \cr & {\text{Use the quotient rule}} \cr & y'' = - \frac{{\left( {1 + 2x{y^{1/2}}} \right)\left( {\frac{6}{2}} \right)\left( {{y^{1/2}}} \right)y' - 2{y^{3/2}}\left( {2{y^{1/2}} + 2\left( {\frac{x}{2}} \right){y^{ - 1/2}}y'} \right)}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} \cr & y'' = - \frac{{\left( {1 + 2x{y^{1/2}}} \right)\left( {3{y^{1/2}}} \right)y' - 2{y^{3/2}}\left( {2{y^{1/2}} + x{y^{ - 1/2}}y'} \right)}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} \cr & {\text{Simplifying}} \cr & y'' = - \frac{{\left( {1 + 2x{y^{1/2}}} \right)\left( {3{y^{1/2}}} \right)y' - 2y\left( {2y + xy'} \right)}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} \cr & y'' = - \frac{{\left( {1 + 2x{y^{1/2}}} \right)\left( {3{y^{1/2}}} \right)y'}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} + \frac{{2y\left( {2y + xy'} \right)}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} \cr & y'' = - \frac{{3{y^{1/2}}}}{{1 + 2x{y^{1/2}}}}y' + \frac{{4{y^2}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} + \frac{{2xy}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}}y' \cr & {\text{Replace }}y' = - \frac{{2{y^{3/2}}}}{{1 + 2x{y^{1/2}}}} \cr & y'' = - \frac{{3{y^{1/2}}}}{{1 + 2x{y^{1/2}}}}\left( { - \frac{{2{y^{3/2}}}}{{1 + 2x{y^{1/2}}}}} \right) + \frac{{4{y^2}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} + \frac{{2xy}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}}y' \cr & y'' = \frac{{6{y^2}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} + \frac{{4{y^2}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} + \frac{{2xy}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}}\left( { - \frac{{2{y^{3/2}}}}{{1 + 2x{y^{1/2}}}}} \right) \cr & y'' = \frac{{10{y^2}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^2}}} - \frac{{4x{y^{5/2}}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^3}}} \cr & {\text{Simplifying}} \cr & y'' = \frac{{10{y^2}\left( {1 + 2x{y^{1/2}}} \right) - 4x{y^{5/2}}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^3}}} \cr & y'' = \frac{{10{y^2} + 20x{y^{5/2}} - 4x{y^{5/2}}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^3}}} \cr & y'' = \frac{{10{y^2} + 16x{y^{5/2}}}}{{{{\left( {1 + 2x{y^{1/2}}} \right)}^3}}} \cr & y'' = \frac{{2{y^2}\left( {5 + 8x\sqrt y } \right)}}{{{{\left( {1 + 2x\sqrt y } \right)}^3}}} \cr} $$