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 - Page 161: 65


$f'(x)=\dfrac{x^{2}+2x-7}{(x+1)^{2}}$ $f''(x)=\dfrac{16}{(x+1)^{3}}$

Work Step by Step

$f(x)=\dfrac{x^{2}-7x}{x+1}$ Use the quotient rule to evaluate the first derivative of the function given: $f'(x)=\dfrac{(x+1)(x^{2}-7x)'-(x^{2}-7x)(x+1)'}{(x+1)^{2}}=...$ Evaluate the derivatives indicated in the numerator: $...=\dfrac{(x+1)(2x-7)-(x^{2}-7x)(1)}{(x+1)^{2}}=...$ Evaluate the products indicated in the numerator and simplify: $...=\dfrac{2x^{2}-5x-7-x^{2}+7x}{(x+1)^{2}}=\dfrac{x^{2}+2x-7}{(x+1)^{2}}$ The expression above is the first derivative. Apply the quotient rule again to evaluate the second derivative: $f''(x)=\dfrac{(x+1)^{2}(x^{2}+2x-7)'-(x^{2}+2x-7)[(x+1)^{2}]'}{(x+1)^{4}}=...$ $...=\dfrac{(x+1)^{2}(2x+2)-2(x^{2}+2x-7)(x+1)}{(x+1)^{4}}=...$ Simplify: $...=\dfrac{2x^{3}+6x^{2}+6x+2-2x^{3}-6x^{2}+10x+14}{(x+1)^{4}}=...$ $...=\dfrac{16x+16}{(x+1)^{4}}=\dfrac{16(x+1)}{(x+1)^{4}}=\dfrac{16}{(x+1)^{3}}$ The expression above is the second derivative
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