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: 72

Answer

The equation of the tangent line is $y=\dfrac{2x+16}{25}.$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=4x; u'(x)=4$ $v(x)=x^2+6; v'(x)=2x$ $f'(x)=\frac{4(x^2+6)-(4x)(2x)}{(x^2+6)^2}=\frac{24-4x^2}{(x^2+6)^2}$ $f'(2)=\frac{24-4(2)^2}{((2^2)+6)^2}=\frac{2}{25}$. Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-\frac{4}{5})=\frac{2}{25}(x-2) \rightarrow y=\frac{2x+16}{25}.$
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