Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises: 73

Answer

$y=\dfrac{8x-7}{5}.$

Work Step by Step

Using the Chain Rule: $u=2x^2-7$; $\dfrac{du}{dx}=4x$ $\dfrac{d}{du}f(u)=\dfrac{1}{2\sqrt{u}}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=\dfrac{2x}{\sqrt{2x^2-7}}.$ $f'(4)=\dfrac{2(4)}{\sqrt{2(4)^2-7}}=\dfrac{8}{5}.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-5)=\frac{8}{5}(x-4)\rightarrow y=\dfrac{8x-7}{5}.$ A graphing calculator and a computer algebra system have been used to confirm these results.
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