# How do you use the definition of a derivative to find the derivative of f(x)=-3x^2+2x+9?

Feb 28, 2017

(df_x)/(dx)=color(green)(-6x+2
(see below for method using the definition of derivative)

#### Explanation:

The derivative of a function $f \left(x\right)$ (with respect to $x$) is defined as
$\textcolor{w h i t e}{\text{XXX}} \frac{{\mathrm{df}}_{x}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

Given: $f \left(x\right) = - 3 {x}^{2} + 2 x + 9$

$\textcolor{w h i t e}{\text{XXX}} f \left(x + h\right) = - 3 {\left(x + h\right)}^{2} + 2 \left(x + h\right) + 9$
$\textcolor{w h i t e}{\text{XXXXXXXX}} = - 3 {x}^{2} - 3 x h - 3 {h}^{2} + 2 x + 2 h + 9$

{: (f(x+h)," = ",,-3x^2,-6xh,-3h^2,+2x,+2h,+9,), (-f(x)," = " ,"- [ ",underline(-3x^2),underline(color(white)(_3xh)),underline(color(white)(-3h^2)),underline(+2x),underline(color(white)(+2h)),underline(+9)," ]"), (f(x+h)-f(x)," = ",,,-6xh,-3h^2,,+2h,,) :}

$\Rightarrow \textcolor{w h i t e}{\text{XX}} \frac{f \left(x + h\right) - f \left(x\right)}{h} = - 6 x - 3 h + 2$
and
$\textcolor{w h i t e}{\text{XXX}} {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h} = - 6 x - 3 \cdot \left(0\right) + 2 = \textcolor{g r e e n}{- 6 x + 2}$