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: 36

Answer

$y=-\sqrt{3}x+4$

Work Step by Step

$\dfrac{d}{dx}(7x^2)-\dfrac{d}{dx}(6\sqrt{3}xy)+\dfrac{d}{dx}(13y^2)-\dfrac{d}{dx}(16)=0\rightarrow$ $14x-6\sqrt{3}y-\dfrac{dy}{dx}(6\sqrt{3}x)+\dfrac{dy}{dx}(26y)=0\rightarrow$ $\dfrac{dy}{dx}=\dfrac{14x-6\sqrt{3}y}{6\sqrt{3}x-26y}=\dfrac{7x-3\sqrt{3}y}{3\sqrt{3}x-13y}.$ At $(\sqrt{3}, 1)\rightarrow\dfrac{dy}{dx}=\dfrac{7(\sqrt{3})-3\sqrt{3}(1)}{3\sqrt{3}(\sqrt{3})-13(1)}=-\sqrt{3}$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-1)=-\sqrt{3}(x-\sqrt{3})\rightarrow y=-\sqrt{3}x+4.$
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