Answer
$$f'(x)=e^x(x^2+2x)$$
$$f''(x)=e^x(x^2+4x+2)$$
$$f'''(x)=e^x(X^2+6x+6)$$
$$f^{(4)}(x)=e^x(x^2+8x+12)$$
$$f^{(5)}(x)=e^x(x^2+10x+20)$$
See work for proof.
Work Step by Step
1. Applying the product rule to f(x), we see that $$f'(x)=(\frac{d}{dx}x^2)e^x+(\frac{d}{dx}e^x)(x^2)$$
$$=2xe^x+x^2e^x$$
$$=e^x(x^2+2x)$$
To get the second derivative, we do the product rule again.$$f''(x)=(\frac{d}{dx}x^2+2x)e^x+(\frac{d}{dx}e^x)(x^2+2x)$$
$$=e^x(2x+2)+e^x(x^2+2x)$$
$$=e^x(x^2+4x+2).$$
Repeat this process 3 more times:
$$f'''(x)=(\frac{d}{dx}x^2+4x+2)e^x+(\frac{d}{dx}e^x)(x^2+4x+2)$$
$$=e^x(2x+4)+e^x(x^2+4x+2)$$
$$=e^x(x^2+6x+6)$$
$$f^{(4)}(x)=(\frac{d}{dx}x^2+6x+6)e^x+(\frac{d}{dx}e^x)(x^2+6x+6)$$
$$=e^x(2x+6)+e^x(x^2+6x+6)$$
$$=e^x(x^2+8x+12)$$
$$f^{(5)}(x)=(\frac{d}{dx}x^2+8x+12)e^x+(\frac{d}{dx}e^x)(x^2+8x+12)$$
$$=e^x(2x+8)+e^x(x^2+8x+12)$$
$$e^x(x^2+10x+20)$$.
By now, we can deduce that $f^{(n)}(x)=e^x(x^2+2nx+n^2-n)$.
Proof:
Base case:
$$f^{(0)}(x)=f(x)=e^xx^2=e^x(x^2+2(0)x+0^2-0)$$
Now Assume that the claim holds for all n-1. Then $f^{(n)}(x)=\frac{d}{dx}f^{(n-1)}(x)=\frac{d}{dx}e^x(x^2+2(n-1)x+(n-1)^2-(n-1)$ by inductive assumption. Then by the product rule, we see that that is equal to $e^x(2x+2(n-1))+e^x(x^2+2(n-1)x+(n-1)^2-n+1)$
$$=e^x(x^2+2(n-1)x+2x+2(n-1)+(n-1)^2-n+1)$$
$$=e^x(x^2+2x(n-1+1)+n^2+2n-n-2+1+1$$
$$=e^x(x^2+2nx+n^2-n)$$$$Q.E.D$$
