Calculus: Early Transcendentals 8th Edition

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