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 - Page 188: 40


$f(x)=(x^2-1)e^x$ $f'(x)=e^x(x^2+2x-1)$ $f''(x)=e^x(x^2+4x+1)$

Work Step by Step

To solve this problem, we need to remember two things: The product rule and the differentiation of an exponential function. Product Rule: $\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$ Differentiation of Exponential Function $\frac{d}{dx}[e^x]=e^x$ So now, you can solve this problem. Knowing that $f(x)=(x^2-1)e^x$ We can use the rules above to find the derivative of $f(x)$. $f'(x)=2x(e^x)+e^x(x^2-1)$ $f'(x)=e^x(2x+x^2-1)$ $f'(x)=e^x(x^2+2x-1)$ And that's your final first derivative. Now, we can do the same thing again to find the second derivative: $f''(x)=e^x(x^2+2x-1)+(2x+2)e^x$ $f''(x)=e^x(x^2+2x-1+2x+2)$ $f''(x)=e^x(x^2+4x+1)$. And that's your final second derivative! So your final answer is: $f(x)=(x^2-1)e^x$ $f'(x)=e^x(x^2+2x-1)$ $f''(x)=e^x(x^2+4x+1)$. You can now check that on a graph to make sure those are reasonable answers.
Small 1504525023
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.