University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 159: 91


a) $dy/dx=1$ b) $dy/dx=1$

Work Step by Step

a) $y=(u/5)+7$ and $u=5x-35$ We have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\Big(\frac{u}{5}+7\Big)'(5x-35)'=\frac{1}{5}\times5=1$$ b) $y=1+(1/u)$ and $u=1/(x-1)$ We have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\Big(1+\frac{1}{u}\Big)'\Big(\frac{1}{x-1}\Big)'=\Big(\frac{-1(u)'}{u^2}\Big)\Big(\frac{-1(x-1)'}{(x-1)^2}\Big)$$ $$\frac{dy}{dx}=\Big(-\frac{1}{u^2}\Big)\Big(-\frac{1}{(x-1)^2}\Big)=\frac{1}{u^2(x-1)^2}$$ Here we substitute $u=1/(x-1)$: $$\frac{dy}{dx}=\frac{1}{\frac{1}{(x-1)^2}\times(x-1)^2}=\frac{1}{1}=1$$ In both cases, we see that the results are the same and equal with $dy/dx$ when $y=x$. So writing the function as a composite in different ways does not change the value of its derivative.
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