Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.1 Definition of the Derivative - Exercises - Page 103: 41


$$ -\frac{1}{16}$$ $$y=-\frac{1}{16} x+\frac{3}{4}$$

Work Step by Step

Let $f(x)=\frac{1}{\sqrt{x}} .$ Then \begin{aligned} f^{\prime}(4) &=\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h}\\ &=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{4+h}}-\frac{1}{2}}{h}\\ &=\lim _{h \rightarrow 0} \frac{\frac{2-\sqrt{4+h}}{2 \sqrt{4+h}} \cdot \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}}{h}\\ &=\lim _{h \rightarrow 0} \frac{\frac{4-4-h}{4 \sqrt{4+h}+2(4+h)}}{h} \\ &=\lim _{h \rightarrow 0} \frac{-1}{4 \sqrt{4+h}+2(4+h)}=-\frac{1}{16} \end{aligned} At $a=4$, the tangent line is \begin{aligned} y&=f^{\prime}(4)(x-4)+f(4)\\ &=-\frac{1}{16}(x-4)+\frac{1}{2}\\ &=-\frac{1}{16} x+\frac{3}{4} \end{aligned}
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