## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 817: 40

#### Answer

$\dfrac{s^5}{2}$ and $\dfrac{5ts^4}{2}$

#### Work Step by Step

Consider $\dfrac{dw}{dt}=\dfrac{dw}{dx}\dfrac{dx}{dt}+\dfrac{d w}{d y}\dfrac{dy}{dt}$ $\implies xys^2+(x^2/2)(-s/t^2)=\dfrac{s^5}{2}$ Now, $\dfrac{dw}{ds}=xy(2ts)+(\dfrac{x^2}{2})(\dfrac{1}{t})$ $\implies (ts^2)(\dfrac{s}{t})(2ts) +(ts^2)^2(2)(1/t)=\dfrac{5ts^4}{2}$

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