Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.1 Exercises - Page 105: 74

Answer

$\frac{x-6}{\left|x-6\right|}$

Work Step by Step

Step 1: Use the chain rule $\frac{df\left(a\right)}{dx}=\frac{df}{da}\cdot \frac{da}{dx}$ $f=\left|a\right|,\:\:a=x-6$ $=\frac{d}{da}\left(\left|a\right|\right)\frac{d}{dx}\left(x-6\right)$ $\frac{d}{da}\left(\left|a\right|\right)=\frac{a}{\left|a\right|}$ Step 2: Use the Sum Difference rule: $\frac{d}{dx}\left(x-6\right)$$=\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(6\right)$ $\frac{d}{dx}\left(x\right)=1$ Step 3: Since the derivative of a constant is zero than: $\frac{d}{dx}\left(6\right)=0$ $=\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(6\right)$ = 1 - 0 = 1 Remember we used $a = x-6$ so substitute in for $a$ $\frac{a}{\left|a\right|}\cdot \:1$ = $\frac{x-6}{\left|x-6\right|}\cdot \:1$ Since anything multiplied by 1 is itself, we can simply forget about the 1 and simplify $\frac{x-6}{\left|x-6\right|}\cdot \:1$ to $\frac{x-6}{\left|x-6\right|}$ Answer: $\frac{x-6}{\left|x-6\right|}$
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