Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises: 52

Answer

$a.$ The derivative is $f'(x)=-\frac{1}{x^2}$. $b.$ The equation is $y=-\frac{1}{25}x-\frac{2}{5}$.

Work Step by Step

$a.$ Using the definition of the derivative at $x$ we have $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to0}\frac{-h}{hx(x+h)}=\lim_{h\to0}\frac{-1}{x(x+h)}=\frac{-1}{x(x+0)}=-\frac{1}{x^2}.$$ $b.$ The equation of the tanget at the point $P(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using this with $a=-5$, $f(a)=f(-1)=1/(-5)=-1/5.$ and $f'(a)=f'(-1)=-1/(-5)^2=-1/25$ we have $$y-\left(-\frac{1}{5}\right)=-\frac{1}{25}(x-(-5))\Rightarrow y+\frac{1}{5}=-\frac{1}{25}x-\frac{1}{5},$$ which gives $$y=-\frac{1}{25}x-\frac{2}{5}.$$
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