Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 69

Answer

$h'(r)=\dfrac{r-6\sqrt{r}-1}{2\sqrt{r}(r+1)^{2}}$

Work Step by Step

$h(r)=\dfrac{2-r-\sqrt{r}}{r+1}$ Rewrite the square root using a rational exponent: $h(r)=\dfrac{2-r-\sqrt{r}}{r+1}=\dfrac{2-r-r^{1/2}}{r+1}$ Evaluate the derivative using the quotient rule: $h'(r)=\dfrac{(r+1)(2-r-r^{1/2})'-(2-r-r^{1/2})(r+1)'}{(r+1)^{2}}=...$ Evaluate the derivatives indicated in the numerator: $...=\dfrac{(r+1)\Big(-1-\dfrac{1}{2}r^{-1/2}\Big)-(2-r-r^{1/2})(1)}{(r+1)^{2}}=...$ Simplify: $...=\dfrac{(r+1)\Big(-1-\dfrac{1}{2r^{1/2}}\Big)-2+r+r^{1/2}}{(r+1)^{2}}=...$ $...=\dfrac{-r-\dfrac{r}{2r^{1/2}}-1-\dfrac{1}{2r^{1/2}}-2+r+r^{1/2}}{(r+1)^{2}}=...$ $...=\dfrac{-\Big(\dfrac{r+1}{2r^{1/2}}\Big)-3+r^{1/2}}{(r+1)^{2}}=\dfrac{\dfrac{-(r+1)-6r^{1/2}+2r}{2r^{1/2}}}{(r+1)^{2}}=...$ $...=\dfrac{-r-1-6r^{1/2}+2r}{2r^{1/2}(r+1)^{2}}=\dfrac{r-6r^{1/2}-1}{2r^{1/2}(r+1)^{2}}=...$ Change the rational exponents to square roots: $...=\dfrac{r-6\sqrt{r}-1}{2\sqrt{r}(r+1)^{2}}$
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