Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=x^2+x`?

Answer 1

`d/(dx) (x^2+x) = 2x+1`

Explanation:

The limit definition of the derivatives states that:

`f'(barx) = lim_(x->barx) frac (f(x)-f(barx)) (x-barx)`

As `f(x) =x^2+x`

`f'(barx) = lim_(x->barx) frac (x^2+x-barx^2-barx) (x-barx)=lim_(x->barx) [frac (x^2-barx^2) (x-barx)+ frac (x-barx) (x-barx)]=lim_(x->barx) frac ((x+barx)(x-barx)) (x-barx)+ 1= lim_(x->barx) (x+barx) + 1 = 2barx +1`

As this holdes for every `x=barx` we have in general that:

`d/(dx) (x^2+x) = 2x+1`