Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 32


$$ f'(x) =4(x+1)(x^2+2x+3)e^{(x^2+2x+3)^2}.$$

Work Step by Step

Recall that $(e^x)'=e^x$ Recall that $(x^n)'=nx^{n-1}$ Since we have $$ f(x)= e^{(x^2+2x+3)^2}$$ then the derivative $ f'(x)$, using the chain rule, is given by $$ f'(x)=e^{(x^2+2x+3)^2}((x^2+2x+3)^2)'=e^{(x^2+2x+3)^2}2(x^2+2x+3)(2x+2)\\ =4(x+1)(x^2+2x+3)e^{(x^2+2x+3)^2}.$$
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