Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 33

Answer

$y=-x+7.$

Work Step by Step

$\dfrac{d}{dx}((y-3)^2)=\dfrac{d}{dx}(4(x-5))\rightarrow$ $\dfrac{dy}{dx}(2(y-3))=4\rightarrow$ $\dfrac{dy}{dx}=\dfrac{2}{y-3}.$ At $(6, 1)\rightarrow\dfrac{dy}{dx}=\dfrac{2}{1-3}=-1.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-1)=-1(x-6)\rightarrow y=-x+7.$
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