Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 36

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{1}{{2\sqrt x }}\sinh \left( {\sqrt x } \right){\sinh ^2}\left( {x{y^2}} \right) + 2{y^2}\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right) \cr & {f_y}\left( {x,y} \right) = 4xy\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right) \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = \cosh \left( {\sqrt x } \right){\sinh ^2}\left( {x{y^2}} \right) \cr & {\text{Calculate the partial derivative}} \cr & {\text{ }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x{\text{ and use the product rule}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {\cosh \left( {\sqrt x } \right)} \right){\sinh ^2}\left( {x{y^2}} \right) + \frac{\partial }{{\partial x}}\left( {{{\sinh }^2}\left( {x{y^2}} \right)} \right)\cosh \left( {\sqrt x } \right) \cr & {\text{Solving the derivatives}} \cr & {f_x}\left( {x,y} \right) = \sinh \left( {\sqrt x } \right)\left( {\frac{1}{{2\sqrt x }}} \right){\sinh ^2}\left( {x{y^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2\sinh \left( {x{y^2}} \right)\frac{\partial }{{\partial x}}\left( {\sinh \left( {x{y^2}} \right)} \right)\cosh \left( {\sqrt x } \right) \cr & {f_x}\left( {x,y} \right) = \frac{{\sinh \left( {\sqrt x } \right){{\sinh }^2}\left( {x{y^2}} \right)}}{{2\sqrt x }} + 2\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right)\left( {{y^2}} \right)\cosh \left( {\sqrt x } \right) \cr & {\text{simplifying}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{2\sqrt x }}\sinh \left( {\sqrt x } \right){\sinh ^2}\left( {x{y^2}} \right) + 2{y^2}\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right) \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y,{\text{ treat }}x{\text{ as a constant}} \cr & {\text{and use the product rule}} \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\cosh \left( {\sqrt x } \right){{\sinh }^2}\left( {x{y^2}} \right)} \right] \cr & {\text{pull out the constant }}\cosh \left( {\sqrt x } \right) \cr & {f_y}\left( {x,y} \right) = \cosh \left( {\sqrt x } \right)\frac{\partial }{{\partial y}}\left[ {{{\sinh }^2}\left( {x{y^2}} \right)} \right] \cr & {\text{use the chain rule}} \cr & {f_y}\left( {x,y} \right) = \cosh \left( {\sqrt x } \right)\left( 2 \right)\left( {\sinh \left( {x{y^2}} \right)} \right)\frac{\partial }{{\partial y}}\left( {\sinh \left( {x{y^2}} \right)} \right) \cr & {f_y}\left( {x,y} \right) = 2\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right)\left( {2xy} \right) \cr & {\text{simplifying}} \cr & {f_y}\left( {x,y} \right) = 4xy\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right) \cr & \cr & {\text{The partial derivatives are:}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{2\sqrt x }}\sinh \left( {\sqrt x } \right){\sinh ^2}\left( {x{y^2}} \right) + 2{y^2}\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right) \cr & {f_y}\left( {x,y} \right) = 4xy\cosh \left( {\sqrt x } \right)\sinh \left( {x{y^2}} \right)\cosh \left( {x{y^2}} \right) \cr} $$