Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 108: 7


$y=x+1$, see figure.

Work Step by Step

Given the curve as $y=2\sqrt x$ and a point $(1,2)$ on the curve, we can find the slope of the tangent line at this point as: $m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to0}\frac{2\sqrt {1+h}-2\sqrt 1}{h}=\lim_{h\to0}\frac{2(\sqrt {1+h}-1)(\sqrt {1+h}+1)}{h(\sqrt {1+h}+1)}=\lim_{h\to0}\frac{2(1+h-1)}{h(\sqrt {1+h}+1)}=\lim_{h\to0}\frac{2}{\sqrt {1+h}+1}=\frac{2}{\sqrt {1+0}+1}=1$ The equation for the tangent line at $(1,2)$ can then be written as $y-2=(x-1)$ or $y=x+1$, as shown in the figure.
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