## Calculus: Early Transcendentals 8th Edition

$$f'(x)=2x-6x^2$$
$$f(x)=x^2-2x^3$$ According to definition, $$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{[(x+h)^2-2(x+h)^3]-(x^2-2x^3)}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{[x^2+h^2+2xh-2x^3-2h^3-6x^2h-6xh^2]-x^2+2x^3}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{h^2+2xh-2h^3-6x^2h-6xh^2}{h}$$ $$f'(x)=\lim\limits_{h\to0}(h+2x-2h^2-6x^2-6xh)$$ $$f'(x)=0+2x-2\times0^2-6x^2-6x\times0$$ $$f'(x)=2x-6x^2$$