Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 28

Answer

$$f'(x)=e^x(\sqrt x+\frac{1}{2\sqrt x})$$ $$f''(x)=e^x(\frac{1}{\sqrt x}+\sqrt x-\frac{1}{4x\sqrt x})$$

Work Step by Step

$$f(x)=\sqrt xe^x$$ $$f(x)=x^{1/2}e^x$$ 1) Find $f'(x)$ Using the Product Rule, we have $$f'(x)=x^{1/2}\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x^{1/2})$$ $$f'(x)=x^{1/2}e^x+e^x\frac{1}{2}x^{-1/2}$$ $$f'(x)=e^x(x^{1/2}+\frac{1}{2}x^{-1/2})$$ $$f'(x)=e^x(\sqrt x+\frac{1}{2\sqrt x})$$ 2) Find $f''(x)$ We would start with $$f'(x)=e^x(x^{1/2}+\frac{1}{2}x^{-1/2})$$ Using the Product Rule again, we have $$f''(x)=e^x\frac{d}{dx}(x^{1/2}+\frac{1}{2}x^{-1/2})+(x^{1/2}+\frac{1}{2}x^{-1/2})\frac{d}{dx}(e^x)$$ $$f''(x)=e^x(\frac{1}{2}x^{-1/2}+\frac{1}{2}\times(-\frac{1}{2}x^{-3/2}))+(x^{1/2}+\frac{1}{2}x^{-1/2})e^x$$ $$f''(x)=e^x(\frac{1}{2}x^{-1/2}-\frac{1}{4}x^{-3/2})+e^x(x^{1/2}+\frac{1}{2}x^{-1/2})$$ $$f''(x)=e^x(x^{-1/2}+x^{1/2}-\frac{1}{4}x^{-3/2})$$ $$f''(x)=e^x(\frac{1}{\sqrt x}+\sqrt x-\frac{1}{4x\sqrt x})$$
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