Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 914: 40

$$z'\left( t \right) = \frac{{t + 2}}{{t + 1}}$$

$$\eqalign{ & {\text{Let }}z = \ln \left( {x + y} \right),\,\,\,\,\,\,\,\,{\text{Where }}x = t{e^t},{\text{ and }}y = {e^t} \cr & \cr & a.{\text{ Replacing }}x{\text{ and }}y{\text{ to write }}z{\text{ as a function of }}t{\text{ and differentiate}} \cr & z = \ln \left( {t{e^t} + {e^t}} \right) \cr & {\text{Differentiate to obtain }}z'\left( t \right) \cr & z'\left( t \right) = \frac{d}{{dt}}\left[ {\ln \left( {t{e^t} + {e^t}} \right)} \right] \cr & z'\left( t \right) = \frac{{\frac{d}{{dt}}\left[ {t{e^t} + {e^t}} \right]}}{{t{e^t} + {e^t}}} \cr & z'\left( t \right) = \frac{{t{e^t} + {e^t} + {e^t}}}{{t{e^t} + {e^t}}} \cr & z'\left( t \right) = \frac{{t{e^t} + 2{e^t}}}{{t{e^t} + {e^t}}} \cr & z'\left( t \right) = \frac{{t + 2}}{{t + 1}} \cr & \cr & b.{\text{ Using the chain rule }}\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr & {\text{Calculating the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\ln \left( {x + y} \right)} \right] = \frac{1}{{x + y}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\ln \left( {x + y} \right)} \right] = \frac{1}{{x + y}} \cr & {\text{Calculating the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {t{e^t}} \right] = t{e^t} + {e^t} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}} \right] = {e^t} \cr & \cr & {\text{Substitute the derivatives into the formula of the chain rule}} \cr & \frac{{dz}}{{dt}} = \left( {\frac{1}{{x + y}}} \right)\left( {t{e^t} + {e^t}} \right) + \left( {\frac{1}{{x + y}}} \right)\left( {{e^t}} \right) \cr & {\text{Write in terms of }}t \cr & \frac{{dz}}{{dt}} = \left( {\frac{1}{{t{e^t} + {e^t}}}} \right)\left( {t{e^t} + {e^t}} \right) + \left( {\frac{1}{{t{e^t} + {e^t}}}} \right)\left( {{e^t}} \right) \cr & {\text{Simplify}} \cr & \frac{{dz}}{{dt}} = 1 + \frac{1}{{t + 1}} \cr & \frac{{dz}}{{dt}} = \frac{{t + 2}}{{t + 1}} \cr} $$