Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises - Page 984: 22


$\dfrac{\partial T}{\partial p} =0$; $\dfrac{\partial T}{\partial q} =\dfrac{-1}{8}$; and $\dfrac{\partial T}{\partial r} =\dfrac{1}{32}$

Work Step by Step

$\dfrac{\partial T}{\partial p} =0\\\dfrac{\partial T}{\partial q} =\dfrac{ (2 q \sqrt {r}+\sqrt q r)-(\sqrt qr)(2 \sqrt r+\dfrac{r}{2 \sqrt q})}{(2q\sqrt r+\sqrt qr)^2}$; $\dfrac{\partial T}{\partial q} =\dfrac{ -8}{64}=\dfrac{-1}{8}$; (When $p=2; q=1; r=4$) $\dfrac{\partial T}{\partial r} =\dfrac{ (2 q \sqrt {r}+\sqrt q r)(\sqrt q)-(\sqrt qr)(\dfrac{q}{\sqrt r}+\sqrt q)}{(2q\sqrt r+\sqrt qr)^2}$; (When $p=2; q=1; r=4$) Thus, we have $\dfrac{\partial T}{\partial r} =\dfrac{2}{64}=\dfrac{1}{32}$
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