# How do you use the definition of a derivative to find the derivative of f(x)=2x^2-3x+5?

Sep 29, 2017

$f ' \left(x\right) = 4 x - 3$

#### Explanation:

There are a few variations on the limit definition of a derivative that are all equivalent, but the one most likely to be of use here is the following one:

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

To use that you need to evaluate (aka "plug in") both $\left(x + h\right)$ and $x$ into the function $f \left(x\right)$, and attempt to reduce what comes out of that, usually by cancellation, before attempting to evaluate the limit.

For instance:

${\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$
$= {\lim}_{h \to 0} \frac{\left(2 {\left(x + h\right)}^{2} - 3 \left(x + h\right) + 5\right) - \left(2 {x}^{2} - 3 x + 5\right)}{h}$

$= {\lim}_{h \to 0} \frac{\left(2 \left({x}^{2} + 2 x h + {h}^{2}\right) - 3 \left(x + h\right) + 5\right) - \left(2 {x}^{2} - 3 x + 5\right)}{h}$

$= {\lim}_{h \to 0} \frac{\left(2 {x}^{2} + 4 x h + 2 {h}^{2} - 3 x - 3 h + 5\right) - \left(2 {x}^{2} - 3 x + 5\right)}{h}$

$= {\lim}_{h \to 0} \frac{2 {x}^{2} + 4 x h + 2 {h}^{2} - 3 x - 3 h + 5 - 2 {x}^{2} + 3 x - 5}{h}$

$= {\lim}_{h \to 0} \frac{\textcolor{red}{\cancel{2 {x}^{2}}} + 4 x h + 2 {h}^{2} \textcolor{b l u e}{\cancel{- 3 x}} - 3 h \textcolor{g r e e n}{\cancel{+ 5}} \textcolor{red}{\cancel{- 2 {x}^{2}}} \textcolor{b l u e}{\cancel{+ 3 x}} \textcolor{g r e e n}{\cancel{- 5}}}{h}$

$= {\lim}_{h \to 0} \frac{4 x h + 2 {h}^{2} - 3 h}{h} = {\lim}_{h \to 0} \frac{h \left(4 x + 2 h - 3\right)}{h}$

$= {\lim}_{h \to 0} \left(4 x + 2 h - 3\right) = 4 x - 3 \text{ } \left[A\right]$

In the last cancellation step [A] we are able to factor and cancel the $h$ terms from the numerator and denominator because in a limit we are examining when $h$ approaches 0, not when it is 0.

Once you learn basic derivatives in Calculus (or end-year Precalculus), you can verify that this is indeed the derivative of $2 {x}^{2} - 3 x + 5$ based upon the "Power Rule" of derivatives.