Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 966: 84


$L \dfrac{\partial P}{\partial L}+K \dfrac{\partial P}{\partial K} =(\alpha+\beta)P$

Work Step by Step

$ \dfrac{\partial P}{\partial L}=\alpha bL^{\alpha-1} \times K^{\beta}$ and $ \dfrac{\partial P}{\partial K}=\beta bL^{\alpha} \times K^{\beta-1}$ Consider the LHS such as: $(L) \dfrac{\partial P}{\partial L}+(K) \dfrac{\partial P}{\partial K}= L \times (\alpha bL^{\alpha-1}K^{\beta})+K \times (\beta bL^{\alpha}K^{\beta-1})=\alpha b L^{\alpha}K^{\beta}+\beta b L^{\alpha}K^{\beta}=(\alpha+\beta)b L^{\alpha} K^{\beta}$ This implies that $L \dfrac{\partial P}{\partial L}+K \dfrac{\partial P}{\partial K} =(\alpha+\beta)P$
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