Question

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

Answer 1

I found: `f'(x)=2x+1`

Explanation:

The definition tells us that:

`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

Where `h` is a small increment.

With our function we have:

`f'(x)=lim_(h->0)([(x+h)^2+(x+h)]-[(x^2+x)])/h`

Expanding:

`f'(x)=lim_(h->0)(cancel(x^2)+2xh+h^2+cancel(x)+hcancel(-x^2)cancel(-x))/h`

Collect `h` in the numerator:

`f'(x)=lim_(h->0)(cancel(h)[2x+h+1])/cancel(h)`

As `h->0`:

`f'(x)=lim_(h->0)(2x+h+1)=2x+1`