Calculus 8th Edition

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

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 695: 1


$\frac{dy}{dx}$= $\frac{(1+t)^{1.5}}{2}$

Work Step by Step

x=$\frac{t}{1+t}$ Use Quotient Rule $\frac{dx}{dt}$ = $\frac{(1+t) * (t)' - (t) * (1+t)' }{(1+t)^{2}}$ $\frac{dx}{dt}$ = $\frac{1}{(1+t)^2}$ y=$\sqrt {1+t}$ Use Power Rule y=$(1+t)^{1/2}$ $\frac{dy}{dt}$ = $(1/2)(1+t)^{-1/2}$ $\frac{dy}{dt}$ = $\frac{1}{2 \sqrt{1+t} }$ $\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ $\frac{dy}{dx}$ = $\frac{\frac{1}{2 \sqrt{1+t} }}{\frac{1}{(1+t)^2}}$ $\frac{dy}{dx}$ = $\frac{(1+t)^2}{2 \sqrt{1+t}}$ $\frac{dy}{dx}$ = $\frac{(1+t)^{1.5}}{2}$
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