# Using the limit definition, how do you differentiate f(x)=x^2-4x+23?

Mar 3, 2016

$2 x - 4$

#### Explanation:

The limit definition of the derivative is given by:

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

Putting $f \left(x\right)$ into the formula we get:

${\lim}_{h \to 0} \frac{{\left(x + h\right)}^{2} - 4 \left(x + h\right) + 23 - {x}^{2} + 4 x - 23}{h}$

Now expend out the brackets to get:

${\lim}_{h \to 0} \frac{{x}^{2} + 2 x h + {h}^{2} - 4 x - 4 h + 23 - {x}^{2} + 4 x - 23}{h}$

Gathering the like terms:

${\lim}_{h \to 0} \frac{\textcolor{red}{{\cancel{x}}^{2}} + 2 x h + {h}^{2} - \textcolor{b l u e}{\cancel{4 x}} - 4 h + \textcolor{g r e e n}{\cancel{23}} - \textcolor{red}{{\cancel{x}}^{2}} + \textcolor{b l u e}{\cancel{4 x}} - \textcolor{g r e e n}{\cancel{23}}}{h}$

$= {\lim}_{h \to 0} \frac{2 x h - 4 h + {h}^{2}}{h}$

$= {\lim}_{h \to 0} 2 x - 4 + h$

Now evaluate the limit and we are left with:

$= 2 x - 4$