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