## Calculus: Early Transcendentals 8th Edition

$$f'(x)=e^x(x^3+3x^2+1)$$ $$f''(x)=e^x(x^3+6x^2+6x+1)$$
$$f(x)=(x^3+1)e^x$$ 1) Find $f'(x)$ Using the Product Rule, we have $$f'(x)=e^x\frac{d}{dx}(x^3+1)+(x^3+1)\frac{d}{dx}(e^x)$$ $$f'(x)=e^x\times3x^2+(x^3+1)\times e^x$$ $$f'(x)=3e^xx^2+e^xx^3+e^x$$ $$f'(x)=e^x(x^3+3x^2+1)$$ 2) Find $f''(x)$ Also using the Product Rule, we have $$f''(x)=e^x\frac{d}{dx}(x^3+3x^2+1)+(x^3+3x^2+1)\frac{d}{dx}(e^x)$$ $$f''(x)=e^x(3x^2+6x)+e^x(x^3+3x^2+1)$$ $$f''(x)=e^x(x^3+6x^2+6x+1)$$