University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 137: 74

Answer

The detailed explanations are below.

Work Step by Step

a) We examine $\frac{d}{dx}(\frac{1}{v})$: According to the Derivative Quotient Rule, we have: $$\frac{d}{dx}\Big(\frac{1}{v}\Big)=\frac{v\frac{d}{dx}(1)-1\frac{dv}{dx}}{v^2}=\frac{0\times v-\frac{dv}{dx}}{v^2}=-\frac{\frac{dv}{dx}}{v^2}=-\frac{1}{v^2}\frac{dv}{dx}$$ This is in fact the Reciprocal Rule. So the Reciprocal Rule is in essence still the Derivative Quotient Rule for $\frac{d}{dx}(\frac{u}{v})$, yet now with $u=1$. b) We can use the Derivative Product Rule together with the Reciprocal Rule to deduce the Derivative Quotient Rule, as follows: $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{d}{dx}\Big(u\times\frac{1}{v}\Big)$$ Apply the Derivative Product Rule here: $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{du}{dx}\times\frac{1}{v}+\frac{d}{dx}\Big(\frac{1}{v}\Big)\times u$$ Next, we use Reciprocal Rule here to calculate $\frac{d}{dx}(\frac{1}{v})$: $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{du}{dx}\times\frac{1}{v}+\Big(-\frac{1}{v^2}\frac{dv}{dx}\Big)u$$ $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{\frac{du}{dx}}{v}-\frac{u\frac{dv}{dx}}{v^2}$$ $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$ We have thus proved the Derivative Quotient Rule.
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