Question

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

Answer 1

`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)|)))`