Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.6 The Chain Rule - Exercises - Page 808: 9


$\dfrac{\partial h}{\partial t_2}=0$

Work Step by Step

We have $\dfrac{\partial h}{\partial x}=\dfrac{1}{y}; \dfrac{\partial h}{\partial y}=\dfrac{-x}{y^2}; \dfrac{\partial x}{\partial t_2}=t_1; \dfrac{\partial y}{\partial t_2}=t_1^2$ We apply the chain rule to obtain: $\dfrac{\partial h}{\partial t_2}=\dfrac{\partial h}{\partial x} \dfrac{\partial x}{\partial t_2}+\dfrac{\partial h}{\partial y} \dfrac{\partial y}{\partial t_2} \\=t_1(\dfrac{1}{y})-t_1^2 (\dfrac{x}{y^2})$ We are given that $x=t_1t_2; y=t_1^2t_2$ Thus, $\dfrac{\partial h}{\partial t_2}=t_1(\dfrac{1}{t_1^2t_2})-t_1^2 (\dfrac{t_1t_2}{(t_1^2t_2)^2})$ Therefore, we get $\dfrac{\partial h}{\partial t_2}=0$
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