## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 7 - Practice Exercises - Page 421: 21

#### Answer

$\dfrac{1}{e}$ $m/s$

#### Work Step by Step

Here, $\dfrac{dy}{dx}=\dfrac{1}{x}(\dfrac{dx}{dt})$ As we are given that $\dfrac{dx}{dt}=\sqrt x$ Then, we have $\dfrac{dy}{dx}=\dfrac{1}{x}(\sqrt x)=\dfrac{1}{x^{1/2}}$ Now, at$(e^2,2)$ $\dfrac{dy}{dt}=\dfrac{1}{\sqrt{e^{2}}}=\dfrac{1}{e}$

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