# How do I use the limit definition of derivative to find f'(x) for f(x)=5x-9x^2 ?

Sep 19, 2014

Let us find $f ' \left(x\right)$ by using the limit definition.

Start with $f \left(x\right)$.

$f \left(x\right) = 5 x - 9 {x}^{2}$

Let us find $f \left(x + h\right)$.

$f \left(x + h\right) = 5 \left(x + h\right) - 9 {\left(x + h\right)}^{2}$
$= 5 x + 5 h - 9 \left({x}^{2} + 2 x h + {h}^{2}\right)$
$= 5 x + 5 h - 9 {x}^{2} - 18 x h - 9 {h}^{2}$

Let us find the difference quotient.

$\frac{f \left(x + h\right) - f \left(x\right)}{h}$

by plugging in the expression we found above,

$= \frac{5 x + 5 h - 9 {x}^{2} - 18 x h - 9 {h}^{2} - \left(5 x - 9 {x}^{2}\right)}{h}$

by cancelling out $5 x$'s and $- 9 {x}^{2}$'s,

$= \frac{5 h - 18 x h - 9 {h}^{2}}{h}$

by factoring $h$ out in the numerator,

$= \frac{h \left(5 - 18 x - 9 h\right)}{h}$

by cancelling out $h$'s,

$= 5 - 18 x - 9 h$

Now, we can find $f ' \left(x\right)$.

$f ' \left(x\right) = {\lim}_{h \to 0} \left(5 - 18 x - 9 h\right) = 5 - 18 x - 9 \left(0\right) = 5 - 18 x$