Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 6

$$\,\,\,\,\frac{{dz}}{{dt}} = {e^t}\cosh \left( {\frac{{t{e^t}}}{2}} \right)\sinh \left( {\frac{{t{e^t}}}{2}} \right) + t{e^t}\cosh \left( {\frac{{t{e^t}}}{2}} \right)\sinh \left( {\frac{{t{e^t}}}{2}} \right)$$

$$\eqalign{ & z = {\cosh ^2}xy;\,\,\,\,\,\,\,\,\,x = t/2,\,\,\,\,\,\,\,y = {e^t} \cr & {\text{Calculate the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial y}}{{\partial x}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{{\cosh }^2}xy} \right) \cr & \,\,\,\,{\text{treat }}y{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = 2\cosh xy\left( {\sinh xy} \right)\left( y \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = 2y\cosh xy\sinh xy \cr & and \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{{\cosh }^2}xy} \right) \cr & \,\,\,\,{\text{treat }}x{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = 2\cosh xy\left( {\sinh xy} \right)\left( x \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = 2x\cosh xy\sinh xy \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {t/2} \right) = 1/2 \cr & \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{e^t}} \right) = {e^t} \cr & \cr & {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr & \,\,\,\,\,{\text{Substitute the derivatives}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {2y\cosh xy\sinh xy} \right)\left( {\frac{1}{2}} \right) + \left( {2x\cosh xy\sinh xy} \right)\left( {{e^t}} \right) \cr & \,\,\,\,{\text{Where }}x = 1/2{\text{ and }}\,\,\,\,\,y = {e^t} \cr & \,\,\,\,\frac{{dz}}{{dt}} = {e^t}\cosh \left( {\frac{{t{e^t}}}{2}} \right)\sinh \left( {\frac{{t{e^t}}}{2}} \right) + 2\left( {\frac{t}{2}} \right){e^t}\cosh \left( {\frac{{t{e^t}}}{2}} \right)\sinh \left( {\frac{{t{e^t}}}{2}} \right) \cr & \,\,\,\,\,{\text{simplifying}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = {e^t}\cosh \left( {\frac{{t{e^t}}}{2}} \right)\sinh \left( {\frac{{t{e^t}}}{2}} \right) + t{e^t}\cosh \left( {\frac{{t{e^t}}}{2}} \right)\sinh \left( {\frac{{t{e^t}}}{2}} \right) \cr} $$