Calculus 10th Edition

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

Chapter 2 - Differentiation - Review Exercises - Page 158: 55



Work Step by Step

Let the equation be of the form $y=\frac{u}{v}$ Let $u=1$ $\frac{du}{dx}=0$ Let $v=x^{2}+4$ $\frac{dv}{dx}=2x$ The equation for the derivative is: $\frac{v \times \frac{du}{dx} - u\times\frac{dv}{dx}}{v^{2}}$ Using the above values, it can be found that the derivative is equal to: $\frac{(x^{2}+4)\times (0)-(1)\times(2x)}{(x^{2}+4)^{2}}$ $=\frac{-2x}{(x^{2}+4)^{2}}$
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