Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.3 Exercises: 69

Answer

The equation of the tangent line is $y=-\frac{1}{2}x+2$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=8; u'(x)=0$ $v(x)=x^2+4; v'(x)=2x$ $f'(x)=\frac{(0)(x^2+4)-(2x)(8)}{(x^2+4)^2}=-\frac{16x}{(x^2+4)^2}$. $f'(2)=-\frac{32}{64}=-\frac{1}{2}$. Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-1)=-\frac{1}{2}(x-2)\rightarrow y=-\frac{1}{2}x+2$.
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