# How do you differentiate f(x)= (3x+4)(2x-5)?

Mar 9, 2017

$f ' \left(x\right)$ = $12 x$ - 7

#### Explanation:

First multiply out the brackets using FOIL or a different method:

$f \left(x\right)$ = $6 {x}^{2}$- $7 x$ - 20

Then differentiate:

$f ' \left(x\right)$ = $12 x$ - 7

The -20 disappears because there is no $x$ attached to it

( $f ' \left(x\right)$ is the same as saying $\frac{\mathrm{dy}}{\mathrm{dx}}$)

http://www.s-cool.co.uk/a-level/maths/differentiation/revise-it/basic-differentiation

Mar 9, 2017

$f ' \left(x\right) = 12 x - 7$

#### Explanation:

Set $y = \textcolor{b l u e}{\left(3 x + 4\right)} \textcolor{g r e e n}{\left(2 x - 5\right)}$

multiply everything in the right bracket by everything in the left

$y = \textcolor{g r e e n}{\textcolor{b l u e}{3 x} \left(2 x - 5\right) \textcolor{b l u e}{+ 4} \left(2 x - 5\right)}$

$y = 6 {x}^{2} - 7 x - 25$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Shortcut method:}}$

The general case: given that $y = a {x}^{n}$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = a n {x}^{n - 1}$

Note that the differential of any constant is 0

differentiation each term in turn

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(6 {x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(- 7 x\right) + \frac{d}{\mathrm{dx}} \left(- 25\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 x - 7 + 0$

Or the current trend is that you write it like:

$\textcolor{g r e e n}{f ' \left(x\right) = 12 x - 7}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{First principle method (the hard way)}}$

Let $x$ grow the very small amount of $\delta x$
As a consequence $y$ will change a small amount $\delta y$

So $y = 6 {x}^{2} - 7 x - 25 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

becomes

$y + \delta y = 6 {\left(x + \delta x\right)}^{2} - 7 \left(x + \delta x\right) - 25$

$y + \delta y = 6 \left({x}^{2} + 2 x \delta x + {\left(\delta x\right)}^{2}\right) - 7 x - 7 \delta x - 25$

$y + \delta y = 6 {x}^{2} + 12 x \delta x + 6 {\left(\delta x\right)}^{2} - 7 x - 7 \delta x - 25. E q u a t i o n \left(2\right)$

$E q u a t i o n \left(2\right) - E q u a t i o n \left(1\right)$

$y + \delta y = 6 {x}^{2} + 12 x \delta x + 6 {\left(\delta x\right)}^{2} - 7 x - 7 \delta x - 25$
$\underline{y \text{ "=6x^2" "-7x" } - 25}$
$\text{ } \delta y = 0 \textcolor{w h i t e}{{x}^{2}} + 12 x \delta x + 6 {\left(\delta x\right)}^{2} + 0 \textcolor{w h i t e}{x} - 7 \delta x + 0$

Divide both sides by $\delta x$

$\frac{\delta y}{\delta x} = 12 x - 7 + 6 \delta x$

Now consider $\delta x$ as becoming increasingly small so that it is almost but not quite 0. The this equation becomes

${\lim}_{\delta x \to 0} \frac{\delta y}{\delta x} = {\lim}_{\delta x \to 0} \left(12 x - 7 + 6 \delta x\right)$

$\textcolor{g r e e n}{\frac{\mathrm{dy}}{\mathrm{dx}} = 12 x - 7 + 0}$ as above in the shortcut method