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

Jan 29, 2017

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

#### Explanation:

The definition tells us that:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

Where $h$ is a small increment.

With our function we have:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\left[{\left(x + h\right)}^{2} + \left(x + h\right)\right] - \left[\left({x}^{2} + x\right)\right]}{h}$

Expanding:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\cancel{{x}^{2}} + 2 x h + {h}^{2} + \cancel{x} + h \cancel{- {x}^{2}} \cancel{- x}}{h}$

Collect $h$ in the numerator:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\cancel{h} \left[2 x + h + 1\right]}{\cancel{h}}$

As $h \to 0$:

$f ' \left(x\right) = {\lim}_{h \to 0} \left(2 x + h + 1\right) = 2 x + 1$