Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 808: 69


$v_x= \dfrac{\ln v}{\ln u \ln v -1}$

Work Step by Step

We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to $x$. $1=v_x \ln u +(v) (u^{-1} ) u_x$ and $0=u_x \ln v +(u) (v^{-1} ) v_x$ So, $u_x= \dfrac{-(u/v) v_x}{\ln v}$ Now, $1=v_x \ln u +(v) (u^{-1} ) (u_x= \dfrac{-(u/v) v_x}{\ln v})$ $\implies v_x= \dfrac{\ln v}{\ln u \ln v -1}$
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