Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 25

$$\frac{{\partial t}}{{\partial x}} = \frac{{{x^2} + {y^2}}}{{4{x^2}{y^3}}}{\text{ and }}\frac{{\partial t}}{{\partial y}} = \frac{{ - 3{x^2} + {y^2}}}{{4x{y^4}}}$$

$$\eqalign{ & t = u/v;\,\,\,\,\,\,\,\,\,\,\,u = {x^2} - {y^2},\,\,\,\,\,\,\,v = 4x{y^3} \cr & \cr & {\text{Substitute }}{x^2} - {y^2}{\text{ for }}u{\text{ and }}4x{y^3}{\text{ for }}v{\text{ into }}t = u/v \cr & t = \frac{{{x^2} - {y^2}}}{{4x{y^3}}} \cr & {\text{Simplify}} \cr & t = \frac{{{x^2}}}{{4x{y^3}}} - \frac{{{y^2}}}{{4x{y^3}}} \cr & t = \frac{x}{{4{y^3}}} - \frac{1}{{4xy}} \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial t}}{{\partial x}}{\text{ and }}\frac{{\partial t}}{{\partial y}} \cr & \frac{{\partial t}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{x}{{4{y^3}}} - \frac{1}{{4xy}}} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial t}}{{\partial x}} = \frac{1}{{4{y^3}}} - \frac{1}{{4y}}\left( { - \frac{1}{{{x^2}}}} \right) \cr & \frac{{\partial t}}{{\partial x}} = \frac{1}{{4{y^3}}} + \frac{1}{{4{x^2}y}} \cr & \frac{{\partial t}}{{\partial x}} = \frac{{{x^2} + {y^2}}}{{4{x^2}{y^3}}} \cr & and \cr & \frac{{\partial t}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{x}{{4{y^3}}} - \frac{1}{{4xy}}} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial t}}{{\partial y}} = \frac{x}{4}\left( { - 3{y^{ - 4}}} \right) - \frac{1}{{4x}}\left( { - \frac{1}{{{y^2}}}} \right) \cr & \frac{{\partial t}}{{\partial y}} = - \frac{{3x}}{{4{y^4}}} + \frac{1}{{4x{y^2}}} \cr & \frac{{\partial t}}{{\partial y}} = \frac{{ - 3{x^2} + {y^2}}}{{4x{y^4}}} \cr} $$