## Calculus: Early Transcendentals 8th Edition

Both methods come out with the result: $$f'(x)=-8x^3+6x^2-2x+1$$
$$f(x)=(1+2x^2)(x-x^2)$$ 1) Using the Product Rule: $$f'(x)=(x-x^2)\frac{d}{dx}(1+2x^2)+(1+2x^2)\frac{d}{dx}(x-x^2)$$ $$f'(x)=(x-x^2)\times4x+(1+2x^2)\times(1-2x)$$ $$f'(x)=4x^2-4x^3+1-2x+2x^2-4x^3$$ $$f'(x)=-8x^3+6x^2-2x+1$$ 2) Performing the multiplication first: $$f(x)=x-x^2+2x^3-2x^4$$ So, $$f'(x)=\frac{d}{dx}(x-x^2+2x^3-2x^4)$$ $$f'(x)=-8x^3+6x^2-2x+1$$ The answers agree.