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

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Answer 1 · 2017-01-29T23:28:10

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

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 $h->0$ :

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

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