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

1 Answer
Jan 29, 2017

Answer:

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#