Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 926: 83


$\dfrac{\partial R}{\partial R_{1}}=[\dfrac{R}{R_{1}}]^2$

Work Step by Step

We need to apply $\dfrac{\partial}{\partial R_{1}}$ to both sides of the given equation. Apply chain rule in the LHS equation as R is a function of $R_{1}.$ and in the RHS $R_{2},$ and $R_{3}$ have to be treated as constants. $\dfrac{\partial}{\partial R_{1}}[R^{-1}]=\dfrac{\partial}{\partial R_{1}}[R_{1}^{-1}]+0+0$ This implies that $-R^{-2}\times \dfrac{\partial R}{\partial R_{1}}=-R_{1}^{-2}$ Hence, we have $\dfrac{\partial R}{\partial R_{1}}=[\dfrac{R}{R_{1}}]^2$
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