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: 37

Answer

$y=\dfrac{\sqrt{3}x+16\sqrt{3}}{6}.$

Work Step by Step

$\dfrac{d}{dx}(x^2y^2)-\dfrac{d}{dx}(9x^2)-\dfrac{d}{dx}(4y^2)=0\rightarrow$ $(2y^2x+\dfrac{dy}{dx}(2x^2y))-18x-\dfrac{dy}{dx}(8y)=0\rightarrow$ $y^2x-9x=\dfrac{dy}{dx}(4y-x^2y)\rightarrow$ $\dfrac{dy}{dx}=\dfrac{y^2x-9x}{4y-x^2y}.$ At $(-4, 2\sqrt{3})\rightarrow\dfrac{dy}{dx}=\dfrac{(2\sqrt{3})^2(-4)-9(-4)}{4(2\sqrt{3})-(-4)^2(2\sqrt{3})}=\dfrac{\sqrt{3}}{6}.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-2\sqrt{3})=\dfrac{\sqrt{3}}{6}(x+4)\rightarrow y=\dfrac{\sqrt{3}x+16\sqrt{3}}{6}.$
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