# How do you use the definition of a derivative to find the derivative of f(x)=2x^2+5x?

Apr 27, 2017

$f ' \left(x\right) = 4 x + 5$

#### Explanation:

By definition of the derivative:

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

So with $f \left(x\right) = 2 {x}^{2} + 5 x$ we have;

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\left\{2 {\left(x + h\right)}^{2} + 5 \left(x + h\right)\right\} - \left\{2 {x}^{2} + 5 x\right\}}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \frac{\left\{2 \left({x}^{2} + 2 h x + {h}^{2}\right) + 5 x + 5 h\right\} - \left\{2 {x}^{2} + 5 x\right\}}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \frac{\left\{2 {x}^{2} + 4 h x + 2 {h}^{2} + 5 x + 5 h\right\} - \left\{2 {x}^{2} + 5 x\right\}}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \frac{2 {x}^{2} + 4 h x + 2 {h}^{2} + 5 x + 5 h - 2 {x}^{2} - 5 x}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \frac{4 h x + 2 {h}^{2} + 5 h}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \left(4 x + 2 h + 5\right)$
$\text{ } = 4 x + 5$