Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 141: 57


Equation of tangent: $x+2y-5=0$ Equation of normal: $2x-y=0$

Work Step by Step

$y=\frac{3x+1}{x^2+1}$. Differentiating with respect to $x$ at $(1,2)$, $$\frac{dy}{dx}=\frac{3(x^2+1)-(3x+1)(2x)}{(x^2+1)^2}=\frac{-3x^2-2x+3}{(x^2+1)^2}$$ $$\frac{dy}{dx}\bigg|_{x=1}=\frac{-3(1)^2-2(1)+3}{(1^2+1)^2}=-\frac{1}{2}$$ Thus, the slope of the required tangent line is $-1/2$. Thus, the equation of tangent line (using point-slope form), $$y-2=-\frac{1}{2}(x-1)$$ $$2y-4=-x+1\implies x+2y-5=0.$$ If the slope of the tangent is $m_1=-\frac{1}{2}$ then the slope of normal $m_2$ can found out by the relation, $m_1 \times m_2 = -1$. Thus, $m_2=2$. Using point-slope form, $$y-2=2(x-1)$$ $$y-2=2x-2\implies 2x-y=0.$$
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