Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 74


$$\frac{d^2}{dx^2}(x^2+9)^5 =10(x^2+9)^4+80x^2(x^2+9)^3.$$

Work Step by Step

Recall that $(x^n)'=nx^{n-1}$ We have $$\frac{d}{dx}(x^2+9)^5=5(x^2+9)^4(2x)=10x(x^2+9)^4$$ and hence $\frac{d^2}{dx^2}(x^2+9)^5=10\frac{d}{dx} x(x^2+9)^4$ Use the product rule: $=10(x^2+9)^4+40x(x^2+9)^3(2x)$ $=10(x^2+9)^4+80x^2(x^2+9)^3$
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