University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 202: 80



Work Step by Step

$$q=(5p^2+2p)^{-3/2}$$ To find $dp/dq$, we would use the methods of implicit differentiation. Differentiate both sides of the equation with respect to $q$: $$1=-\frac{3}{2}(5p^2+2p)^{-5/2}(5p^2+2p)'$$ $$1=-\frac{3}{2}(5p^2+2p)^{-5/2}\Big(10p\frac{dp}{dq}+2\frac{dp}{dq}\Big)$$ $$1=-\frac{3}{2}(5p^2+2p)^{-5/2}(5p+1)2\frac{dp}{dq}$$ $$1=-3(5p^2+2p)^{-5/2}(5p+1)\frac{dp}{dq}$$ Then calculate for $dp/dq$: $$\frac{dp}{dq}=-\frac{1}{3(5p^2+2p)^{-5/2}(5p+1)}$$ $$\frac{dp}{dq}=-\frac{(5p^2+2p)^{5/2}}{3(5p+1)}$$
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