Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 24

$$\frac{{dR}}{{d\phi }} = 5{e^{5\phi }}$$

$$\eqalign{ & R = {e^{2s - {t^2}}};\,\,\,\,\,\,\,\,s = 3\phi ,\,\,\,\,\,\,\,t = {\phi ^{1/2}} \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial R}}{{\partial s}}{\text{ and }}\frac{{\partial R}}{{\partial t}} \cr & \,\,\,\,\frac{{\partial R}}{{\partial s}} = \frac{\partial }{{\partial s}}\left( {{e^{2s - {t^2}}}} \right) \cr & \,\,\,\,{\text{treat }}t{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial R}}{{\partial s}} = {e^{2s - {t^2}}}\frac{\partial }{{\partial s}}\left( {2s - {t^2}} \right) \cr & \,\,\,\,\frac{{\partial R}}{{\partial s}} = 2{e^{2s - {t^2}}} \cr & \cr & \,\,\,\,\frac{{\partial R}}{{\partial t}} = \frac{\partial }{{\partial t}}\left( {{e^{2s - {t^2}}}} \right) \cr & \,\,\,\,{\text{treat }}s{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial R}}{{\partial t}} = {e^{2s - {t^2}}}\frac{\partial }{{\partial t}}\left( {2s - {t^2}} \right) \cr & \,\,\,\,\frac{{\partial R}}{{\partial t}} = - 2t{e^{2s - {t^2}}} \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{ds}}{{d\phi }}{\text{ and }}\frac{{dt}}{{d\phi }} \cr & \,\,\,\,\frac{{ds}}{{d\phi }} = \frac{d}{{d\phi }}\left( {3\phi } \right) = 3 \cr & \,\,\,\,\frac{{dt}}{{d\phi }} = \frac{d}{{d\phi }}\left( {{\phi ^{1/2}}} \right) = \frac{1}{2}{\phi ^{ - 1/2}} \cr & \cr & {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr & \,\,\,\,\frac{{dR}}{{d\phi }} = \frac{{\partial R}}{{\partial s}}\frac{{ds}}{{d\phi }} + \frac{{\partial R}}{{\partial t}}\frac{{dt}}{{d\phi }} \cr & \,\,\,\,\,{\text{Substitute the derivatives}} \cr & \,\,\,\,\frac{{dR}}{{d\phi }} = \left( {2{e^{2s - {t^2}}}} \right)\left( 3 \right) + \left( { - 2t{e^{2s - {t^2}}}} \right)\left( {\frac{1}{2}{\phi ^{ - 1/2}}} \right) \cr & \,\,\,\,{\text{Where }}\,s = 3\phi ,\,\,\,\,\,\,\,t = {\phi ^{1/2}} \cr & \,\,\,\,\frac{{dR}}{{d\phi }} = 6{e^{2\left( {3\phi } \right) - {{\left( {{\phi ^{1/2}}} \right)}^2}}} + \left( { - 2{\phi ^{1/2}}{e^{2\left( {3\phi } \right) - {{\left( {{\phi ^{1/2}}} \right)}^2}}}} \right)\left( {\frac{1}{2}{\phi ^{ - 1/2}}} \right) \cr & \,\,\,\,\,{\text{simplifying}} \cr & \,\,\,\,\frac{{dR}}{{d\phi }} = 6{e^{5\phi }} - {e^{5\phi }} \cr & \,\,\,\,\frac{{dR}}{{d\phi }} = 5{e^{5\phi }} \cr} $$