Using the limit definition, how do you differentiate f(x)=2x^2-x?

1 Answer
May 14, 2016

f'(x)=4x-1

Explanation:

The limit definition of a derivative states that

f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h

Substituting f(x)=2x^2-x into f'(x),

f'(x)=lim_(hrarr0)((2(x+h)^2-(x+h))-(2x^2-x))/h

From this point on, you want to expand and simplify.

f'(x)=lim_(hrarr0)((2x^2+4xh+2h^2-x-h)-2x^2+x)/h

f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)(2x^2)+4xh+2h^2color(blue)cancelcolor(black)(-x)-hcolor(red)cancelcolor(black)(-2x^2)color(blue)cancelcolor(black)(+x))/h

f'(x)=lim_(hrarr0)(4xh+2h^2-h)/h

Factor out h from the numerator.

f'(x)=lim_(hrarr0)(h(4x+2h-1))/h

f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)h(4x+2h-1))/color(red)cancelcolor(black)h

f'(x)=lim_(hrarr0)4x+2h-1

Plugging in h=0,

f'(x)=lim_(hrarr0)4x+2(0)-1

f'(x)=lim_(hrarr0)4x-1

color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=4x-1)color(white)(a/a)|)))