Question

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

Answer 1

`f'(x)=4x-1`

Explanation:

The limit definition of the derivative states that the derivative of the function `f(x)` is

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

Here, `f(x)=2x^2-x`, so `f(x+h)=2(x+h)^2-(x+h)`.

The function's derivative is

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

Distribute and simplify.

`=lim_(hrarr0)([2(x^2+2xh+h^2)-x-h]-2x^2+x)/h`

`=lim_(hrarr0)([2x^2+4xh+2h^2-x-h]-2x^2+x)/h`

`=lim_(hrarr0)(color(red)(cancel(color(black)(2x^2)))+4xh+2h^2color(red)(cancel(color(black)(-x)))-hcolor(red)(cancel(color(black)(-2x^2)))color(red)(cancel(color(black)(+x))))/h`

`=lim_(hrarr0)(4xh+2h^2-h)/h`

Divide an `h` from each term.

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

Now the limit can be evaluated by plugging in `0` for `h`.

`f'(x)=4xcolor(red)(cancel(color(black)(+2(0))))-1`

`f'(x)=4x-1`