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: 71

Answer

The equation of the tangent line is $y=\frac{12x-16}{25}.$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=16x; u'(x)=16$ $v(x)=x^2+16; v'(x)=2x$ $f'(x)=\frac{(16)(x^2+16)-(16x)(2x)}{(x^2+16)^2}=-\frac{16(x^2-16)}{(x^2+16)^2}$ $f'(-2)=-\frac{16((-2)^2-16)}{((-2)^2+16)^2}=\frac{12}{25}$. Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y+\frac{8}{5})=\frac{12}{25}(x+2)\rightarrow y=\frac{12x-16}{25}.$
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